POJ3061 Subsequence

标签：poj3061

Description

Input

Output

Sample Input

2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5

Sample Output

2
3

Source

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define maxn 100010

int S[maxn], N, K;

int main() {
    int i, T, cnt, tmp;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &N, &K);
        for(i = 0; i < N; ++i) {
            scanf("%d", &S[i]);
            if(i) S[i] += S[i - 1];
        }
        if(S[N - 1] < K) {
            printf("0\n");
            continue;
        }
        cnt = N + 1;
        for(i = 0; i < N && S[i] + K <= S[N - 1]; ++i) {
            tmp = std::lower_bound(S + i, S + N, S[i] + K) - S - i;
            cnt = std::min(cnt, tmp);
        }
        printf("%d\n", cnt);
    }
    return 0;
}

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define maxn 100010

int S[maxn], N, K;

int main() {
    int i, T, cnt, sum, s, t;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &N, &K);
        for(i = 0; i < N; ++i) 
            scanf("%d", &S[i]);
        cnt = N + 1; 
        sum = s = t = 0;
        for(i = 0; i < N; ++i) {
            sum += S[i];
            while(sum >= K) {
                t = i - s + 1;
                cnt = std::min(cnt, t);
                sum -= S[s++];
            }
        }
        if(cnt > N) cnt = 0;
        printf("%d\n", cnt);
    }
    return 0;
}

POJ3061 Subsequence

标签：poj3061

原文地址：http://blog.csdn.net/chang_mu/article/details/40860081