AtCoder Beginner Contest 247 C D E题解
作者：互联网
https://atcoder.jp/contests/abc247/tasks/abc247_c
递归即可解决：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>
#define lson 2*x
#define rson 2*x+1
long long qupower(int a, int b) {
long long ans = 1;
while (b) {
if (b & 1)ans = ans * a;
b >>= 1;
a = a * a;
}
return ans;
}
inline int read() {
int an = 0, x = 1; char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') {
x = -1;
}
c = getchar();
}
while (c >= '0'&&c <= '9') {
an = an * 10 + c - '0'; c = getchar();
}
return an * x;
}
const int N = 65540;
int a[N];
void sol(int x){
if(x==1){printf("1 ");return;}
sol(x-1);
printf("%d ",x);
sol(x-1);
}
int main(){
//ios::sync_with_stdio(false);
int n; n = read();
sol(n);
}
https://atcoder.jp/contests/abc247/tasks/abc247_d
不能用queue来模拟，复杂度过不去。我们只需要记录每一段的位置（前缀和维护），以及该段的value。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>
#define lson 2*x
#define rson 2*x+1
long long qupower(int a, int b) {
long long ans = 1;
while (b) {
if (b & 1)ans = ans * a;
b >>= 1;
a = a * a;
}
return ans;
}
inline int read() {
int an = 0, x = 1; char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') {
x = -1;
}
c = getchar();
}
while (c >= '0'&&c <= '9') {
an = an * 10 + c - '0'; c = getchar();
}
return an * x;
}
int Q;
//queue<ll> q;
ll ans = 0;
const int N = 2e5+5;
ll v[N],pos[N];
ll cnt=1;
ll cur_p = 1;
ll real_p = 0;
int main(){
//ios::sync_with_stdio(false);
Q = read();
while(Q--){
int q; q = read();
if(q==1){
ll x,c; scanf("%lld %lld", &x,&c);
pos[cnt] = pos[cnt-1]+c;
v[cnt] = x;
cnt+=1;
}
else{
ll c; scanf("%lld", &c);
ll next_p = real_p+c;
//cout<<"next pos:"<<next_p<<endl;
ans=0;
for(;cur_p<cnt;cur_p++){
if(pos[cur_p]<=next_p){
ans+= ll(v[cur_p]*(pos[cur_p]-real_p));
real_p = pos[cur_p];
}
else{
ans+= ll(v[cur_p]*(next_p-real_p));
real_p = next_p;
break;
}
}
//cout<<real_p<<endl;
printf("%lld\n", ans);
}
}
}
https://atcoder.jp/contests/abc247/tasks/abc247_e
给定一个序列，以及两个数 X,Y. 问有多少子序列（连续的）满足：该序列的最小值为Y，最大值为X。
注意到的一点是：如果有一个数字不满足条件，即 a>X or a<Y，此时包含该数字的序列都不满足，所以我们可以根据不满足条件的数字来分割成小串。现在问题回到了如何对于满足条件的序列来统计数量。
我们使用双指针 L,R. 开始的时候移动右指针 R。当找到最小值和最大值时，即可break来统计数量。具体来说，如果区间 [ L, W ]满足条件，那么区间 [ L, i ]，W< i < R都满足条件。最后移动左指针。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 200005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>
#define lson 2*x
#define rson 2*x+1
long long qupower(int a, int b) {
long long ans = 1;
while (b) {
if (b & 1)ans = ans * a;
b >>= 1;
a = a * a;
}
return ans;
}
inline int read() {
int an = 0, x = 1; char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') {
x = -1;
}
c = getchar();
}
while (c >= '0'&&c <= '9') {
an = an * 10 + c - '0'; c = getchar();
}
return an * x;
}
const int N = 2e5 + 5;
int a[N];
int X,Y;
int n;
ll res=0;
ll cal(vector<int> &vec){
ll ans=0;
int l = 0, r = vec.size()-1;
int lp=l,rp=l;
int cnt_max=0, cnt_min = 0;
//int len = r-l+1;
while(lp<=r){
while(rp<=r && (cnt_max==0 || cnt_min==0)){
if(vec[rp]==X)cnt_max+=1;
if(vec[rp]==Y)cnt_min+=1;
rp+=1;
}
if(cnt_max>0 && cnt_min>0){
ans = ans+ll(r-rp+2);
}
if(vec[lp]==X)cnt_max-=1;
if(vec[lp]==Y)cnt_min-=1;
lp+=1;
}
return ans;
}
void sol(){
int i=0;
int st=0,ed=0;
while(i<n){
// split to subproblem
vector<int> vec;
while(i<n && a[i]<=X && a[i]>=Y){vec.push_back(a[i]);i+=1;}
if(vec.size()>0){
res+= cal(vec);
}
else i+=1;
}
}
int main(){
//ios::sync_with_stdio(false);
n = read();
X = read(); Y = read();
for(int i=0;i<n;i++) a[i] = read();
sol();
printf("%lld\n", res);
}
标签：AtCoder,int,题解,ll,long,247,ans,include,define 来源： https://www.cnblogs.com/xinyu04/p/16193107.html