ZOJ

题意：给定N个数，求和大于等于S的最短连续子序列的长度。

分析：滑动窗口即可。两种写法。

1、

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN];
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int N, S;
        scanf("%d%d", &N, &S);
        int tot = 0;
        for(int i = 0; i < N; ++i){
            scanf("%d", &a[i]);
            tot += a[i];
        }
        if(tot < S){
            printf("0
");
            continue;
        }
        int st = 0, et = 0;
        int sum = 0;
        int cnt = 0x7f7f7f7f;
        while(et < N){
            while(et < N && sum + a[et] < S){
                sum += a[et];
                ++et;
            }
            cnt = min(cnt, et - st + 1);
            sum -= a[st];
            ++st;
        }
        printf("%d
", cnt);
    }
    return 0;
}

2、

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN];
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int N, S;
        scanf("%d%d", &N, &S);
        int tot = 0;
        for(int i = 0; i < N; ++i){
            scanf("%d", &a[i]);
            tot += a[i];
        }
        if(tot < S){
            printf("0
");
            continue;
        }
        int st = 0, et = 0;
        int sum = 0;
        int cnt = 0x7f7f7f7f;
        for(int i = 0; i < N; ++i){
            sum += a[i];
            while(sum - a[st] >= S){
                sum -= a[st];
                ++st;
            }
            if(sum >= S){
                cnt = min(cnt, i - st + 1);
                sum -= a[st];
                ++st;
            }
        }
        printf("%d
", cnt);
    }
    return 0;
}