2021中国大学生程序设计竞赛部分题解（CCPC）- 网络选拔赛（重赛）
作者：互联网
文章目录
- 一、1002 Kanade Doesn't Want to Learn CG
- 二、1004 Primality Test
- 三、1005 Monopoly
- 四、1006 Nun Heh Heh Aaaaaaaaaaa
- 五、1010 Bigraph Extension
- 六、1011 Jumping Monkey
一、1002 Kanade Doesn’t Want to Learn CG
简单数学问题：判断位置即可
#include <bits/stdc++.h>
#define MaxN 0x3f3f3f3f
#define MinN 0xc0c0c0c0
#define bug(x) cout << #x"=" << x << endl;
#define BUG(a,b) cout << #a"=" << a << " " << #b"=" << b << endl;
#define ALL(a) a.begin(), a.end()
#define rep1(i,a,b) for(int i=(a);i<=(b);i++)
#define per1(i,a,b) for(int i=(a);i>=(b);i--)
#define rep2(i,a,b) for(int i=(a);i<(b);i++)
#define per2(i,a,b) for(int i=(a);i>(b);i--)
#define mset(a,b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define EPS 1e-9
#define MAX(a,b) (((a)>(b)? (a):(b)))
#define MIN(a,b) (((a)>(b)? (b):(a)))
#define buff ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define INF 0x3f3f3f3f
#define pb push_back
#define fi first
#define se second
#define mp make_pair
#define pf(x) printf("%d\n",x);
#define sf(x) scanf("%d",&x);
using namespace std;
void solve(){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
double x0,x1,y0,y1,y2;
scanf("%lf%lf%lf%lf%lf",&x0,&x1,&y0,&y1,&y2);
double x = 0.0;
x = (-sqrt(b*b-4*a*(c-y0))-b)/(2*a);
// 这里注意a是负数：取哪个根的时候要抉择一下.
// cout << "x=" << x << " x0=" << x0 << " x1=" << x1 << endl;
if(x>x0 && x1>x) puts("YES");
else if(x<=x0) puts("NO");
else{
double yy = a*x1*x1+b*x1+c;
if(yy<=y2){
if(2*x1-x>x0 && 2*x1-x<x1){
puts("YES");
}else{
puts("NO");
}
}else{
puts("NO");
}
}
}
int main()
{
//buff;
int t;
scanf("%d",&t);
while(t--){
solve();
}
return 0;
}
二、1004 Primality Test
签到题。
通过水几组样例可以发现规律：当 x>1 时输出NO，当x=1时输出YES.
正经分析：通过观察发现，f(x)和f(f(x))是连续的质数，质数一定是奇数，故g(x)一定是整数.（从这里就可以看出2比较特殊（因为2是偶数））。假设g(x)是个质数，则g(x)一定在f(x)和f(f(x))之间，与观察矛盾，故g(x)不可能是质数。
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
long long x;
while(t--){
scanf("%lld",&x);
if(x==1) puts("NO");
else puts("YES");
}
return 0;
}
三、1005 Monopoly
维护一种数据结构…(kungeyyds)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll p = 998244353;
ll a[100010];
struct node {
ll u, x, pos;
friend bool operator<(node s1, node s2) {
if(s1.u == s2.u) return s1.x < s2.x;
return s1.u < s2.u;
}
} q[100010];
#define sangkun
int main() {
#ifdef sangkun
freopen("in.txt", "r", stdin);
#endif
int t;
cin >> t;
while(t--) {
int n, m;
cin >> n >> m;
ll sum = 0;
int op = 0;
map<ll, map<ll, bool>>vis;
for(int i = 1; i <= n; i++) {
scanf("%lld", &a[i]);
sum += a[i];
}
if(sum == 0) {
map<ll, int>ans;
for(int i = 1; i <= n; i++) {
a[i] += a[i - 1];
if(!ans[a[i]]) ans[a[i]] = i;
}
while(m--) {
ll x;
cin >> x;
if(x == 0) {
printf("0\n");
} else if(ans[x]) {
printf("%d\n", ans[x]);
} else printf("-1\n");
}
continue;
}
int ok=0;
if(sum < 0) {
for(int i = 1; i <= n; i++) {
a[i] = -a[i];
}
sum = -sum;
ok=1;
}
q[++op] = {0, 0, 0};
vis[0][0] = 1;
for(int i = 1; i <= n; i++) {
a[i] += a[i - 1];
ll u = abs(a[i]) % sum;
ll x = a[i] / sum;
if(a[i] < 0) u = sum - u, x--;
if(vis[u][x]) continue;
vis[u][x] = 1;
q[++op] = {u, x, i};
}
sort(q + 1, q + 1 + op);
// for(int i=1;i<=op;i++) cout<<q[i].u<<" "<<q[i].x<<" "<<q[i].pos<<endl;
while(m--) {
ll x;
cin >> x;
if(ok) x=-x;
ll u = abs(x) % sum, y = x / sum;
if(x < 0) u = sum - u, y--;
ll l = 1, r = op, ansl = op + 1, ansr = 0, ans = 0;
while(l <= r) {
int mid = (l + r) >> 1;
if(q[mid].u < u) l = mid + 1;
else r = mid - 1, ansl = mid;
}
l = 1, r = op;
while(l <= r) {
int mid = (l + r) >> 1;
if(q[mid].u <= u) l = mid + 1, ansr = mid;
else r = mid - 1;
}
if(q[ansl].u != u) {
printf("-1\n");
continue;
}
l = ansl, r = ansr;
while(l <= r) {
int mid = (l + r) >> 1;
if(q[mid].x <= y) l = mid + 1, ans = mid;
else r = mid - 1;
}
if(ans == 0) {
printf("-1\n");
continue;
}
ans = (y - q[ans].x) * n + q[ans].pos;
cout << ans << endl;
}
}
return 0;
}
四、1006 Nun Heh Heh Aaaaaaaaaaa
DP问题：定义dp(i,j) 表示前i个数中匹配到第j位的方案数.
dp(i,j) = dp(i-1,j) + (if(匹配) +1);
dp[i][j] = (dp[i-1][j] + dp[i-1][j-1]*(s[i]==str[j])) % MOD;
dp(i,9)*后缀a的贡献数即为答案.
后缀a的方案数：2的n次方-1（由二项式定理推出）
处理2的n次方：使用快速幂或者使用左移运算符(1<<i);
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <string>
#define bug(x) cout << #x"=" << x << endl;
#define BUG(a,b) cout << #a"=" << a << " " << #b"=" << b << endl;
using namespace std;
typedef long long ll;
const int N = 100010;
const int MOD = 998244353;
ll mod_power(ll a,ll n,ll mod){ll res = 1;while(n){if(n&1) res = res*a%mod;n >>= 1;a = a*a%mod;}return (res-1);}
ll dp[N][10];
char s[N];
void solve(){
char str[] = " nunhehheh";
ll a[N];
scanf("%s",s+1);
int len = strlen(s+1);
// cout << len << endl;
a[len+1] = 0;
for(int i=len;i>=1;i--){
a[i] = a[i+1] + (s[i]=='a');
}
ll ans = 0;
for(int i=1;i<=len;i++){
for(int j=1;j<=9;j++){
if(j==1) dp[i][j] = dp[i-1][j] + (s[i]==str[j]);
else if(j==9) dp[i][j] = dp[i][j-1]*(s[i]==str[j]); // 这里出现问题.
else dp[i][j] = dp[i-1][j] + dp[i-1][j-1]*(s[i]==str[j]);
}
if(dp[i][9]){
ans = (ans + (dp[i][9]*mod_power(2,a[i+1],MOD))%MOD)%MOD;
// ans = (ans + (dp[i][9]%MOD*((1<<a[i+1])-1)%MOD)%MOD)%MOD; 使用左移运算符计算.
}
}
printf("%lld\n",ans);
}
int main()
{
int t;
scanf("%d",&t);
while(t--){
solve();
}
return 0;
}
五、1010 Bigraph Extension
(kungeyyds)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll p = 998244353;
int f[1010][2];
int pre[2010];
int tmp(int x) {
if(x == pre[x]) return x;
return pre[x] = tmp(pre[x]);
}
int n, m;
bool add(int x, int y) {
y += n;
int A = tmp(x);
int B = tmp(y);
if(A == B) return false;
pre[A] = B;
return true;
}
int main() {
int t;
cin >> t;
while(t--) {
cin >> n >> m;
map<int, map<int, int>>vis;
vector<pair<int, int>>ans;
set<int>s;
for(int i = 1; i <= n; i++) {
pre[i] = i;
pre[i + n] = i+n;
f[i][0] = 0;
f[i][1] = 0;
s.insert(i);
}
for(int i = 1; i <= m; i++) {
int u, v;
scanf("%d%d", &u, &v);
add(u, v);
f[u][0]++;
f[v][1]++;
if(f[v][1] == 2) s.erase(v);
}
if(m == 0) {
f[1][0]++;
f[1][1]++;
add(1, 1);
ans.push_back({1, 1});
}
for(int i = 1; i <= n; i++) {
while(f[i][0] != 2 && ans.size() + m != 2 * n - 1) {
f[i][0]++;
for(auto j : s) {
// cout << tmp(i) << ":" << tmp(j) << endl;
if(!add(i, j)) continue;
f[j][1]++;
ans.push_back({i, j});
// cout << i << "==" << j << endl;
if(f[j][1] == 2)s.erase(j);
break;
}
}
}
std::cout << q0 << ' '
int q1 = 0, q2 = 0;
for(int i = 1; i <= n; i++) {
if(f[i][0] == 1) q1 = i;
if(f[i][1] == 1) q2 = i;
}
ans.push_back({q1, q2});
cout << ans.size() << endl;
for(auto i : ans) printf("%d %d\n", i.first, i.second);
}
return 0;
}
六、1011 Jumping Monkey
先排序然后从小到大枚举;每次把当前点作为并查集的根;然后遍历它的儿子 ;只考虑权值比它小的;那么儿子所在的集的根能到的第一个点就是当前点;这样去求每个点能到的第一个比它大的点;然后倒序去求答案;答案就是能到的第一个点的答案+1.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll p = 998244353;
int u[100100], v[100100];
int a[100100], b[100100];
vector<int>g[100100];
int ans[100010];
int pre[100010];
int tmp(int x) {
if(x == pre[x]) return x;
return pre[x] = tmp(pre[x]);
}
int main() {
int t;
cin >> t;
while(t--) {
int n;
scanf("%d", &n);
for(int i = 1; i < n; i++) {
scanf("%d%d", &u[i], &v[i]);
}
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
b[i] = a[i];
g[i].clear();
ans[i] = 0;
pre[i] = i;
}
sort(b + 1, b + 1 + n);
for(int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;
for(int i = 1; i <= n; i++) {
u[i] = a[u[i]];
v[i] = a[v[i]];
g[u[i]].push_back(v[i]);
g[v[i]].push_back(u[i]);
}
for(int i = 1; i <= n; i++) {
for(auto j : g[i]) {
if(j < i) {
int A = tmp(j);
if(A == i) continue;
pre[A] = i;
ans[A] = i;
}
}
}
for(int i = n; i >= 1; i--) {
ans[i] = ans[ans[i]];
ans[i]++;
}
for(int i = 1; i <= n; i++) printf("%d\n", ans[a[i]]);
}
}
标签：int,题解,ll,CCPC,--,2021,ans,dp,define 来源： https://blog.csdn.net/m0_50435987/article/details/120700577