uva1121 Subsequence(入门级)

给定一个长为N的序列和s，求一个最小的子串使得sum(子序列)>= s;

思路：Twopoint，r维护加入，l维护减除，r-l+1就是这个结果，还要判断下是否存在这样的串。

int a[100001];int n, s;int main(int argc, const char * argv[]){        // freopen("in.txt","r",stdin);    // freopen("out.txt","w",stdout);    while(scanf("%d%d",&n,&s) == 2)    {        int sum = 0;        int tail = 0;        int len = INF;        for (int i = 1;i <= n;++i)        {            scanf("%d", &a[i]);            sum += a[i];            while(sum >= s && tail < i)            {                sum -= a[++tail];                len = min(len, i - tail + 1);            }        }        if (len == INF) cout << 0 << endl;        else cout << len << endl;    }    return 0;}