尺取法——POJ3061

#include <iostream>   //nlogn复杂度的写法
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long  ll;
const int INF = 0x3f3f3f3f;
const int moder = 1e9 + 7;
const int maxn = 100005;

int a[maxn];

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n,s;
        scanf("%d%d",&n,&s);
        for(int i=1;i <= n;i++)
            scanf("%d",&a[i]);

        for(int i=2;i <= n;i++)
        {
            a[i] = a[i] + a[i-1];
        }
        if(a[n] < s) printf("0
");
        else
        {
            int res = n;
            for(int i=1;a[n]-a[i] >= s;i++)
            {
                int t = lower_bound(a+1,a+n+1,s+a[i]) - a; // - a 非 - a - 1
                res = min(res,t-i);
            }
            printf("%d
",res);
        }
    }
    return 0;
}
/*2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5 */

 O(n)复杂度的写法

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long  ll;
const int INF = 0x3f3f3f3f;
const int moder = 1e9 + 7;
const int maxn = 100005;

int a[maxn];

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n,s;
        scanf("%d%d",&n,&s);
        for(int i=1;i <= n;i++)
            scanf("%d",&a[i]);
        int i=1,j=1;
        int sum=0;
        int res = n+1;
        for(;;)
        {
            while(i <= n && sum < s)
            {
                sum += a[i];
                i++;
            }
            if(sum < s) break;
            res = min(res,i-j);
            sum = sum - a[j];
            j++;
        }
        if(res > n) res = 0;
        printf("%d
",res);
    }
    return 0;
}
/*2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5 */

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