poj 3061 尺取法或二分

经典尺取法，复杂度O(n)。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int INF = 999999;
const int N = 100000;
int a[N];

int main ()
{
    int t;
    scanf("%d", &t);
    while ( t-- )
    {
        int n, s;
        scanf("%d%d", &n, &s);
        for ( int i = 0; i < n; i++ ) scanf("%d", a + i);
        int p = 0, q = 0, sum = 0, len = INF;
        while ( 1 )
        {
            while ( q < n && sum < s )
            {
                sum += a[q++];
            }
            if ( sum < s ) break;
            len = min( len, q - p );
            sum -= a[p++];
        }
        if ( len == INF ) len = 0;
        printf("%d
", len);
    }
    return 0;
}

二分也可以做，复杂度多了个log。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

typedef long long ll;
const int INF = 999999;
const int N = 100001;
ll sum[N];

int main ()
{
    int t;
    scanf("%d", &t);
    while ( t-- )
    {
        int n, s;
        scanf("%d%d", &n, &s);
        sum[0] = 0;
        for ( int i = 1; i <= n; i++ )
        {
            scanf("%lld", sum + i);
            sum[i] += sum[i - 1];
        }
        int len = INF;
        for ( int i = n; i > 0; i-- )
        {
            if ( sum[i] < s ) break;
            int pos = upper_bound( sum, sum + i, sum[i] - s ) - sum;
            if ( pos != i )
            {
                len = min( len, i - pos + 1 );
            }
        }
        if ( len == INF ) len = 0;
        printf("%d
", len);
    }
    return 0;
}