CSUST 集训队选拔赛题解
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A题
素数筛 + 线段树（树状数组）
先用素数筛打表，然后线段树更新，遍历求出值，O（1）查询即可
AC代码：
/*num数组 是把记录 数是否存在 存在即为1。
总共有N个数，如何判断第i+1个数到最后一个
数之间有多少个数小于第i个数呢？不妨假设
有一个区间 [1,N]，只需要判断区间[i+1,N]之
间有多少个数小于第i个数。如果我们把总区间初
始化为0，然后把第i个数之前出现过的数都在相应
的区间把它的值定为1，那么问题就转换成了[i+1,N]
值的总和
*/
#include <stdio.h>
#include <bits/stdc++.h>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define size 1000000 + 7
#define min(a, b) a > b ? b : a
typedef long long LL;
#define MAXN 1000005
#define MAXL 1299710
LL prime[MAXN];
LL check[MAXL];
LL PP[MAXL];
void init(){
LL tot = 0;
memset(check, 0, sizeof(check));
for (LL i = 2; i < MAXL; ++i)
{
if (!check[i])
{
prime[tot++] = i;
}
for (LL j = 0; j < tot; ++j)
{
if (i * prime[j] > MAXL)
{
break;
}
check[i*prime[j]] = prime[j];
if (i % prime[j] == 0)
{
break;
}
}
}
}
LL num[size << 2], x[size];
void pushup(LL rt)
{
num[rt] = num[rt << 1] + num[rt << 1 | 1];
}
void build( LL l, LL r, LL rt)
{
num[rt] = 0;
if( l == r) return;
LL m = (l + r) >> 1;
build(lson);
build(rson);
}
LL qurey( LL L, LL R, LL l, LL r, LL rt)
{
if( L <= l && r <= R)
return num[rt];
LL m = (l + r) >> 1;
LL ans = 0;
if(L <= m) ans+=qurey(L, R, lson);
if(R > m) ans+=qurey(L, R, rson);
return ans;
}
void updata( LL p, LL l, LL r, LL rt)
{
if( l == r)
{
num[rt]++;return;
}
LL m = ( l + r) >> 1;
if( p <= m) updata(p, lson);
else updata(p, rson);
pushup(rt);
}
LL a[MAXN + 7];
int main()
{
LL n;
init();
check[1] = 1;
for(LL i = 2 ; i <= MAXN; i++ ){
if(!check[i]) check[i] = i;
}
LL sum = 0;
build(1, MAXN , 1);
for( LL i= 1; i <= MAXN; i++)
{
LL sum = 0;
//scanf("%lld", &x[i]);
sum = qurey(1, check[i], 1, MAXN , 1);
a[i] = sum;
updata(check[i], 1, MAXN, 1);
}
LL t;
scanf("%lld",&t);
while(t--){
LL x;
scanf("%lld",&x);
printf("%lld\n",a[x]);
}
return 0;
}
View Code
B题
并查集
没有蘑菇的建树，求出有多少个没有蘑菇的路，然后求出有蘑菇的路就行
AC代码：
/**
/*@author Victor
/*language C++
*/
//#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
//#include<map>
#include<set>
//#define DEBUG
#define RI register ll
using namespace std;
typedef long long ll;
//typedef __ll128 lll;
const ll N=100000+10;
const ll MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const ll INF = 0x3f3f3f3f;
#define pii pair<ll,ll>
#define pll pair<ll,ll>
#define pil pair<ll , ll>
#define pli pair<ll,ll>
#define pdl pair<double,ll>
#define pld pair<ll,double>
#define pdd pair<double,double>
#define iput(n) scanf("%lld",&n);
#define iiput(a,n) scanf("%lld%lld",&a,&n);
#define iiiput(a,b,c) scanf("%lld%lld%lld",&a,&b,&c);
#define dput(n) scanf("%lf",&n);
#define llput(n) scanf("%lld",&n);
#define cput(n) scanf("%s",n);
#define puti(n) prllf("%lld\n",n);
#define putll(n) prllf("%lld\n",n);
#define putd(n) prllf("%lfd\n",n);
#define _cls(n) memset(n,0,sizeof(n));
#define __cls(n) memset(n,INF,sizeof(n));
//priority_queue <ll,vector<ll>,greater<ll> > Q;//优先队列递增
//priority_queue<ll>Q;//递减
//map<ll,ll>mp;
//set<ll>st;
//stack<>st;
//queue<>Q;
/***********************************************/
//加速输入挂
# define IOS ios::sync_with_stdio(false)
# define FOR(i,a,n) for(ll i=a; i<=n; ++i)
//求二进制中1的个数
//__builtin_popcount(n);
//求2^k
//#define (ll)Pow(2,k) (1LL<<k)
#define to_1(n) __builtin_popcount(n)
//树状数组
#define lowbit(x) (x&-x)
ll fa[333333];
ll pre[333333];
void init(ll n){
for(ll i = 1;i <= n;i++)
fa[i] = i;
}
ll find(ll x) {
if(fa[x] == x)
return x;
return fa[x] = find(fa[x]);
}
void merge(ll x,ll y) {
ll xi = find(x);
ll yi = find(y);
if(xi == yi)
return;
else
fa[xi] = yi;
return ;
}
int main(){
ll n;
IOS;
cin >> n;
init(n);
for(ll i = 1 ; i < n ; i++){
ll u,v,w;
cin >> u >> v >> w;
if(w == 1) continue;
merge(u,v);
}
for(ll i = 1; i <= n;i++){
++pre[find(i)];
}
ll sum = 0;
for(ll i = 1; i <= n;i++){
if(pre[i])
sum += pre[i] * (n - pre[i]);
}
cout << sum << endl;
return 0;
}
View Code
C题
N次最短路
AC代码：
/**
/*@author Victor
/*language C++
*/
#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
using namespace std;
#define MAXN 200010
#define LEN 200010
#define INF 1e9+7
#define MODE 1000000
#define pi acos(-1)
#define g 9.8
typedef long long ll;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
struct edge{
ll next,to,w;
};
edge G[MAXN*2];
ll root;
ll num=0;
ll cnt=0;
ll head[MAXN<<1];
ll depth[MAXN<<1];//深度
ll first[MAXN<<1];//首次出现编号
ll dir[MAXN<<1];//距离
ll que[MAXN<<1];//队列
ll par[MAXN];//并查集父节点
bool vis[MAXN];
ll dp[MAXN][20];
void init(ll n){for(ll i=0;i<=n;i++)par[i]=i;}//初始化并查集
ll _find(ll x){if(par[x]==x)return x;else return par[x]=_find(par[x]);}//查询并查集祖先
void unite(ll x,ll y){x=_find(x),y=_find(y);if(x==y)return;else par[x]=y;}//合并节点
void add(ll u,ll v,ll w){G[num].w=w;G[num].to=v;G[num].next=head[u];head[u]=num++;}//前向星建图
void dfs(ll u,ll dep)//dfs建图
{
vis[u]=true;
que[++cnt]=u;first[u]=cnt;depth[cnt]=dep;
for(ll k=head[u];k!=-1;k=G[k].next)
{
ll v=G[k].to,w=G[k].w;
if(!vis[v]){
dir[v]=dir[u]+w;
dfs(v,dep+1);
que[++cnt]=u;depth[cnt]=dep;
}
}
}
//ST表求LCA
void ST(ll n)
{
for(ll i=1;i<=n;i++)
dp[i][0] = i;
for(ll j=1;(1<<j)<=n;j++)
{
for(ll i=1;i+(1<<j)-1<=n;i++)
{
ll a = dp[i][j-1] , b = dp[i+(1<<(j-1))][j-1];
dp[i][j] = depth[a]<depth[b]?a:b;
}
}
}
//两个节点之间的路径深度最小的就是LCA
ll RMQ(ll l,ll r)
{
ll k=0;
while((1<<(k+1))<=r-l+1)
k++;
ll a = dp[l][k], b = dp[r-(1<<k)+1][k]; //保存的是编号
return depth[a]<depth[b]?a:b;
}
ll LCA(ll u ,ll v)
{
ll x = first[u] , y = first[v];
if(x > y) swap(x,y);
ll res = RMQ(x,y);
return que[res];
}
ll n,m,c;
int main()
{
scanf("%lld",&n);
m = n - 1;
num=0,cnt=0;
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
init(n);
for(ll i=0;i<m;i++)
{
ll u,v,w;
scanf("%lld%lld%lld",&u,&v,&w);
add(u,v,w);
add(v,u,w);
unite(u,v);
}
ll u1,v1,w1;
scanf("%lld%lld%lld",&u1,&v1,&w1);
for(ll i=1;i<=n;i++)
{
if(_find(i)==i){
dir[i]=0;
dfs(i,1);
}
}
ST(2*n-1);
scanf("%lld",&c);
while(c--)
{
ll u,v;
scanf("%lld%lld",&u,&v);
ll lca=LCA(u,v);
ll mid1 = dir[u]+dir[v]-2*dir[lca];
ll lca1=LCA(u,u1);
ll lca2 = LCA(v1,v);
ll mid2 = dir[u]+dir[u1]-2*dir[lca1] + dir[v]+dir[v1]-2*dir[lca2] + w1;
ll lca3=LCA(u,v1); ll lca4 = LCA(u1,v);
ll mid3 = dir[u]+dir[v1]-2*dir[lca3] + dir[v]+dir[u1]-2*dir[lca4] + w1;
ll mine;
if(mid1 > mid2){
mine = mid2;
}
else mine = mid1;
if(mine > mid3){
mine = mid3;
}
//else mine = mid;
printf("%lld\n",mine);
}
return 0;
}
View Code
D题
gcd求最简分数
AC代码：
// .*/@@@]]``...
// .,@@ ,[[\@@@@]`.. .**.*]]]]/@@@@@@@\]]]]*.**.
// *@@ ,OOOO]` .[@@@]]@@@@@@[[[[`* *,[[[\@@@@@@]]`.. .*....*****.....
// .=@^ OOOOOOOOO\ @@@[` .[\@@@\]/@@@@@@@@@@@@@@@@@\`.
// .=@^,OOOOOOOOO[` [@@ *]]]]]]. =@^.
// .@@*=OOOOOO/[` \OOOOOOOOOOO^ =@^.
// .@@ /OOOOO/* ,,OOOOOOO^ @@*
// .@@.OOOO/. ,OOOOOOO^=@^.
// .=@^\OO` =O/OOO @@`.
// .=@^ ,OOO*,@/.
// *@@` [. @@*.
// *@@. @@@^.
// .=@^ =@^.
// *@@ ,]. .@@*
// .=@^ =/ [\@\ .\@^.
// ./@` */@\.
// *@@ *^@@*
// *@@ ]]` ,]]. *^@@.
// .,@/ @@@@@@^ =@@@@@@ ,^@@.
// .@@ =@@@@@@@ @@@@@@@^ =^@@.
// .=@^ @@@@@@^ =@@@@@@ *o^@@*
// .=@^ ,ooooo^ [[` [[[[[` ,[[. ./oooo^ /o=@/.
// .=@^ ,ooooo^ *oooooo. =o/@@`.
// *@@ ** =oo=@/.
// .,@\ /oo\@@.
// .,@@ ,oo//@/.
// .*\@\ ,oo/\@@`.
// .,\@@] ]]]]] .]oo[]@@@`.
// ..,\@@@@]]]* *@@@OOO@@@@\. .,]]]@@@@@/`*.
// .*.*[[[@@@@@/ @@O@@@OO@@@O@@` \@@@[*...
// ./@^ =@OOOO@@@@OOOO@@^ *o\@^.
// .=@/ ,@@@@@@@@@@@@O@O@@OOO@@^,]@@^ =^@@`.
// *@@ ,@@OO@@@@@O@OOO@@@[. *o/@@.
// .=@^ /@@O@@@O@@@OOO@@ ]. \o\@^.
// .@@ ,]]@@@@@@OO@@@@O@@OOO@@\]]]/@@/ =o/@@*
// .,@/ ,[[* \@@OOO@OOOOOOO@@[[[` *o\@@^.
// .=@^ \@@OOOOOOOOO@@@\` oo=@\.
// .@@. .[@@@@@OOOOOO@@` =o^@@*
// *@@ *@@OO@@@@@@ =o^@@.
// *@@ ,@@@[ *` =o^@@^.
// .*.................................................
//
/**
/*@author Victor
/*language C++
*/
//#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
//#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pil pair<int , ll>
#define pli pair<ll,int>
#define pdl pair<double,ll>
#define pld pair<ll,double>
#define pdd pair<double,double>
#define iput(n) scanf("%d",&n);
#define iiput(a,n) scanf("%d%d",&a,&n);
#define iiiput(a,b,c) scanf("%d%d%d",&a,&b,&c);
#define dput(n) scanf("%lf",&n);
#define llput(n) scanf("%lld",&n);
#define cput(n) scanf("%s",n);
#define puti(n) printf("%d\n",n);
#define putll(n) printf("%lld\n",n);
#define putd(n) printf("%lfd\n",n);
#define _cls(n) memset(n,0,sizeof(n));
#define __cls(n) memset(n,INF,sizeof(n));
//priority_queue <int,vector<int>,greater<int> > Q;//优先队列递增
//priority_queue<int>Q;//递减
//map<ll,ll>mp;
//set<ll>st;
//stack<>st;
//queue<>Q;
/***********************************************/
//加速输入挂
# define IOS ios::sync_with_stdio(false)
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
//求二进制中1的个数
//__builtin_popcount(n);
//求2^k
//#define (ll)Pow(2,k) (1LL<<k)
#define to_1(n) __builtin_popcount(n)
//树状数组
#define lowbit(x) (x&-x)
ll gcd(ll a,ll b){
return b?gcd(b,a%b):a;
}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
ll exgcd(ll l,ll r,ll &x,ll &y){if(r==0){x=1;y=0;return l;}else{ll d=exgcd(r,l%r,y,x);y-=l/r*x;return d;}}
//求a关于m的乘法逆元
ll mod_inverse(ll a,ll m){ll x,y;if(exgcd(a,m,x,y)==1)/*ax+my=1*/return (x%m+m)%m;return -1;/*不存在*/}
//快速乘
ll qmul(ll a,ll b,ll m){ll ans=0;ll k=a;ll f=1;/*f是用来存负号的*/if(k<0){f=-1;k=-k;}if(b<0){f*=-1;b=-b;}while(b){if(b&1)ans=(ans+k)%m;k=(k+k)%m;b>>=1;}return ans*f;}
//中国剩余定理CRT (x=ai mod mi)
ll china(ll n, ll *a,ll *m) {ll M=1,y,x=0,d;for(ll i = 1; i <= n; i++) M *= m[i];for(ll i = 1; i <= n; i++) {ll w = M /m[i]; exgcd(m[i], w, d, y);/*m[i]*d+w*y=1*/ x = (x + y*w*a[i]) % M;}return (x+M)%M;}
//筛素数，全局：int cnt,prime[N],p[N];
//void isprime(){cnt = 0;memset(prime,true,sizeof(prime));for(int i=2; i<N; i++){if(prime[i]){p[cnt++] = i;for(int j=i+i; j<N; j+=i)prime[j] = false;}}}
//快速求逆元
//void inverse(){inv[1] = 1;for(int i=2;i<N;i++){if(i >= M) break;inv[i] = (M-M/i)*inv[M%i]%M;}}
//组合数取模，n和m 10^5时，预处理出逆元和阶乘
/*
ll fac[N]={1,1},inv[N]={1,1},f[N]={1,1};
ll C(ll a,ll b){
if(b>a)return 0;
return fac[a]*inv[b]%M*inv[a-b]%M;
}
void init(){//快速计算阶乘的逆元
for(int i=2;i<N;i++){
fac[i]=fac[i-1]*i%M;
f[i]=(M-M/i)*f[M%i]%M;
inv[i]=inv[i-1]*f[i]%M;
}
}
*/
int main(){
int t;
cin >> t;
while(t--){
ll a,b,c,d;
cin >> a >> b >> c >> d;
ll sum =( b - a + 1) * (d - c + 1);
ll su;
if(a > d || c > b) su = 0;
else if(a > c)
{
if(b > d)
su = d - a + 1;
else su = b - a + 1;
}
else if( a <= c){
if(b > d)
su = d - c + 1;
else su = b - c + 1;
}
if(su == 0) printf("0/1\n");
else {
if(sum % su == 0){
printf("1/%lld\n",sum / su);
}
else {
printf("%lld/%lld\n",su / gcd(su,sum) , sum / gcd(su,sum));
}
}
}
return 0;
}
View Code
E题
暴力最小遍历
AC代码：
/**
/*@author Victor
/*language C++
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[1000],b[1000][1000];
int main(){
ll n,k;
scanf("%lld%lld",&n,&k);
ll min = 0;
for(int j = 0 ; j < n ; j++){
for(int i = 0 ; i < k ; i++ ){
scanf("%lld",&a[i]);
}
sort(a,a+k);
for(int z = 0 ; z < k; z++ )
{
b[j][z] = a[z];
// cout << a[z] << endl;
}
}
for(int i = 0 ; i < k ; i++){
for(int j = 0 ; j < n - 1 ; j++){
min += abs(b[j][i] - b[j + 1][i]);
// cout << min << endl;
}
}
printf("%lld\n",min);
return 0;
}
View Code
F题
枚举每点的最短路求和求最小值
AC代码：
/**
/*@author Victor
/*language C++
*/
#include<iostream>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<map>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<fstream>
using namespace std;
typedef pair<int,int> PII;
const int MAXN=1e4+5;
const int INF=0x3f3f3f3f;
bool vis[MAXN];
int dist[MAXN],head[MAXN],tot;
int pre[MAXN];
struct Edge
{
int from,to,cost,nxt;
Edge(){}
Edge(int _from,int _to,int _cost):from(_from),to(_to),cost(_cost){}
}e[MAXN*2];
void addedge(int u,int v,int w)
{
e[tot].from=u;e[tot].to=v;e[tot].cost=w;
e[tot].nxt=head[u];head[u]=tot++;
}
struct qnode
{
int c,v;
qnode(int _c=0,int _v=0):c(_c),v(_v){}
bool operator < (const qnode &rhs) const {return c>rhs.c;}
};
void Dijkstra(int n,int st)//点的编号从1开始
{
memset(vis,false,sizeof(vis));
for(int i=0;i<=n;i++) dist[i]=INF;
priority_queue<qnode> pq;
while(!pq.empty()) pq.pop();
dist[st]=0;
pq.push(qnode(0,st));
qnode frt;
while(!pq.empty())
{
frt=pq.top(); pq.pop();
int u=frt.v;
if(vis[u]) continue;
vis[u]=true;
for(int i=head[u];i!=-1;i=e[i].nxt)
{
int to=e[i].to;
int cost=e[i].cost;
if(dist[to]>dist[u]+cost)
{
dist[to]=dist[u]+cost;
pre[to]=u;
pq.push(qnode(dist[to],to));
}
}
}
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0) break;
tot=0;memset(head,-1,sizeof(head));
int u,v,w;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);addedge(v,u,w);
}
int st,ed;
int sum = 0;
int mine = INF;
for(int st = 1; st <= n;st++){
Dijkstra(n,st);
for(int i = 1 ; i <= n;i++ ) {
sum += dist[i];
}
if(sum < mine){
ed = st;
mine = sum ;
}
sum = 0;
}
printf("%d %d\n",mine , ed);
}
return 0;
}
View Code
G题
逆序对 + 线段树（归并排序
AC代码：
/**
/*@author Victor
/*language C++
*/
#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
#define _cls(a) memset(a,0,sizeof(a))
typedef long long ll;
using namespace std;
ll n, a[200005], tmpA[200005], cnt = 0;
ll p[200005];
void merge_sort(ll l, ll r, ll *A) {
if (l >= r) return ;
ll mid = (l + r) >> 1;
merge_sort(l, mid, A);
merge_sort(mid + 1, r, A);
ll pl = l, pr = mid + 1, tmpp = 0;
while(pl <= mid && pr <= r) {
if (A[pl] <= A[pr]) tmpA[tmpp++] = A[pl++];
else tmpA[tmpp++] = A[pr++], cnt += mid - pl + 1;
}
while(pl <= mid) tmpA[tmpp++] = A[pl++];
while(pr <= r) tmpA[tmpp++] = A[pr++];
for (ll i = 0; i < tmpp; i++) A[i + l] = tmpA[i];
}
int main() {
ll t;
scanf("%lld",&t);
while(t--){
_cls(p);
_cls(a);
_cls(tmpA);
cnt = 0;
scanf("%lld", &n);
for (ll i = 1; i <= n; i++){
ll x;
scanf("%lld", &x);
p[x] = i;
}
for (ll i = 1; i <= n; i++) {
ll x;
scanf("%lld", &x);
a[i] = p[x];
}
merge_sort(1, n, a);
printf("%lld\n", cnt);
}
return 0;
}
View Code
H题
二分 + 前缀和
AC代码：
/**
/*@author Victor
/*language C++
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct name
{
ll w, v;
};
struct name1
{
ll a, b;
};
ll s[422222];
ll o[422222];
name p[400000 + 10];
name1 p1[400000 + 10];
int main(){
ll n,m ,S;
scanf("%lld%lld%lld",&n,&m,&S);
for(ll i = 1 ; i <= n;i ++ )
{
ll w,v;
scanf("%lld%lld",&p[i].w,&p[i].v);
}
ll l = 1, r = 10000000;
ll W ;
for(ll i = 1 ; i <= m; i++){
scanf("%lld %lld",&p1[i].a,&p1[i].b);
}
ll sum = 2000000000000;
for(ll k = 1 ; k <= 100; k++){
W = (l + r) / 2;
memset(s,0,sizeof(s));
memset(o,0,sizeof(o));
for(ll j = 1; j <= n; j++ ){
if(W <= p[j].w){
s[j] = s[j - 1] + p[j].v;
o[j] = 1 + o[j - 1];
}
else {
s[j] = s[j - 1] ;
o[j] = o[j - 1] ;
}
}
ll mine = 0;
ll H = 0;
// ll sum = 2000000;
for(ll i = 1; i <= m ; i++ ){
H += (s[p1[i].b] - s[p1[i].a - 1]) * (o[p1[i].b] - o[p1[i].a - 1]);
}
if(H < S) {
r = W ;
}
else l = W;
mine = abs(H - S);
if(sum > mine) sum = mine;
}
printf("%lld\n",sum);
return 0;
}
View Code
I题
完全背包 + 快速幂
AC代码：
/**
/*@author Victor
/*language C++
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll t;
ll dp[200000 + 7];
ll qpow(ll a,ll b,ll m){
ll ans=1;
ll k=a;
while(b){
if(b&1)ans=ans*k%m;
k=k*k%m;
b>>=1;
}
return ans;
}int main(){
scanf("%lld",&t);
while(t -- ){
memset(dp,0,sizeof(dp));
ll n, m;
dp[1] = 1;
// dp[0] = 1;
scanf("%lld%lld",&n,&m);
if(m==1) dp[n] = qpow(2,n,100000000 + 7);
else{
for(int i = 0;i < m;i++){
dp[i] = 1;
}
for(int i = m ; i <= n;i++){
dp[i] += dp[i - 1] + dp[i - m];
dp[i] %= (100000000 + 7);
}
}
// for(int i = 0 ; i <= n ; i++ ){
// printf("%d\n",dp[i]);
// }
//printf("%d",dp[n]);
printf("%lld\n",dp[n] % (100000000 + 7));
}
return 0;
}
View Code
J题
线段树
AC代码：
/**
/*@author Victor
/*language C++
*/
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <vector>
#define LL long long
using namespace std;
LL INF = 0x3f3f3f3f;
const LL MAX = 1000000 + 50;
LL Sum[MAX << 2];
LL Add[MAX << 2];
LL A[MAX];
//PushUp函数更新节点信息 ，这里是求和
void PushUp(LL rt){
Sum[rt] = Sum[rt<<1] + Sum[rt<< 1|1];
//prLLf("888sum[%ll] = %ll\n", rt, Sum[rt]);
}
//PushDown下推标记
void PushDown(LL rt,LL ln,LL rn){
//ln,rn为左子树，右子树的数字数量。
if(Add[rt]){
//下推标记
Add[rt<<1]+=Add[rt];
Add[rt<<1|1]+=Add[rt];
//修改子节点的Sum使之与对应的Add相对应
Sum[rt<<1]+=Add[rt]*ln;
Sum[rt<<1|1]+=Add[rt]*rn;
//清除本节点标记
Add[rt]=0;
}
}
//Build函数建树
void Build(LL l,LL r,LL rt){ //l,r表示当前节点区间，rt表示当前节点编号
//prLLf("l = %ll r = %ll\n", l, r);
if(l==r) {//若到达叶节点
Sum[rt] = A[l];//储存数组值
//prLLf("A[%ll] = %lld\n", l, A[l]);
//prLLf("sum[%ll] = %ll\n", rt, Sum[rt]);
return;
}
LL m=(l+r)>>1;
//左右递归
Build(l,m,rt<<1);
Build(m+1,r,rt<<1|1);
//更新信息
PushUp(rt);
//prLLf("888sum[%ll] = %ll\n", rt, Sum[rt]);
}
//点修改 A[L] + C
void Update(LL L,LL C,LL l,LL r,LL rt){//l,r表示当前节点区间，rt表示当前节点编号
if(l==r){//到叶节点，修改
Sum[rt] = C;
//prLLf("sum[%ll] = %ll\n", rt, Sum[rt]);
return;
}
LL m=(l+r)>>1;
PushDown(rt,m-l+1,r-m);//下推标记
//根据条件判断往左子树调用还是往右
if(L <= m) Update(L,C,l,m,rt<<1);
else Update(L,C,m+1,r,rt<<1|1);
PushUp(rt);//子节点更新了，所以本节点也需要更新信息
}
//区间修改
void UpdateRange(LL L,LL R,LL C,LL l,LL r,LL rt){//L,R表示操作区间，l,r表示当前节点区间，rt表示当前节点编号
if(L <= l && r <= R){//如果本区间完全在操作区间[L,R]以内
Sum[rt]+=C*(r-l+1);//更新数字和，向上保持正确
Add[rt]+=C;//增加Add标记，表示本区间的Sum正确，子区间的Sum仍需要根据Add的值来调整
return ;
}
LL m=(l+r)>>1;
PushDown(rt,m-l+1,r-m);//下推标记
//这里判断左右子树跟[L,R]有无交集，有交集才递归
if(L <= m) UpdateRange(L,R,C,l,m,rt<<1);
if(R > m) UpdateRange(L,R,C,m+1,r,rt<<1|1);
PushUp(rt);//更新本节点信息
}
//区间查询， 这里是求和
LL QueryRange(LL L,LL R,LL l,LL r,LL rt){//L,R表示操作区间，l,r表示当前节点区间，rt表示当前节点编号
if(L <= l && r <= R){
//在区间内，直接返回
return Sum[rt];
}
LL m=(l+r)>>1;
//下推标记，否则Sum可能不正确
PushDown(rt,m-l+1,r-m);
//累计答案
LL ANS= 0;
if(L <= m) ANS += QueryRange(L,R,l,m,rt<<1);
if(R > m) ANS += QueryRange(L,R,m+1,r,rt<<1|1);
return ANS;
}
LL Query(LL l, LL r, LL rt, LL k){
if(l == r){
return l;
}
LL m = (l + r) >> 1;
if(Sum[(rt << 1)] >= k){
return Query(l, m, rt << 1, k);
} else{
return Query(m + 1, r, rt << 1 | 1, k - Sum[rt << 1]);
}
}
//建树 Build(1,n,1);
//点修改 Update(L,C,1,n,1);
//区间修改 UpdateRange(L,R,C,1,n,1);
//区间查询 LL ANS=Query(L,R,1,n,1);
int main(int argc, char const *argv[])
{
LL n, q;
scanf("%lld%lld", &n, &q);
for(LL i = 1; i <= n; i++){
scanf("%lld", &A[i]);
}
Build(1, n, 1);
while(q--){
char op[3];
LL a , b , c;
scanf("%s",op);
if (op[1] == '1') {
scanf("%lld%lld%lld",&a,&b,&c);
UpdateRange(a,b,c,1,n,1);
} else if(op[1] == '2'){
scanf("%lld%lld%lld",&a,&b,&c);
UpdateRange(a , b , -c , 1 , n , 1);
}
else if(op[1] == '3'){
scanf("%lld%lld",&a,&c);
Update(a,c,1,n,1);
}
else if(op[1] == '4'){
scanf("%lld%lld",&a,&b);
printf("%lld\n",QueryRange(a , b , 1 , n , 1));
}
}
return 0;
}
View Code
标签：lld,int,题解,ll,选拔赛,CSUST,include,LL,define 来源： https://www.cnblogs.com/DWVictor/p/10485829.html