POJ 3061 Subsequence【尺取法】

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原题链接

思路：尺取法，lower_bound

AC代码：

#include <cstdio>  #include <cstring>  #include <algorithm>  #include <iostream>  #include <queue>  using namespace std;int t, n, s, sum[1000002], a[1000002], ans;int main(){    scanf("%d", &t);    while(t--){        scanf("%d %d", &n, &s);        for(int i = 1; i <= n; i++){            scanf("%d", &a[i]);        }        sum[0] = 0;        for(int i = 1; i <= n; i++){            sum[i] = sum[i - 1] + a[i];        }        if(sum[n] < s){            printf("0\n");            continue;        }        ans = n;        for(int i = 1; sum[i] + s <= sum[n]; i++){            int j = lower_bound(sum + i + 1, sum + n + 1, sum[i] + s) - sum;            ans = min(ans, j - i);        }        printf("%d\n", ans);    }    return 0;}

还可以再进行优化，代码为：

#include <cstdio>  #include <cstring>  #include <algorithm>  #include <iostream>  #include <queue>  using namespace std;int t, n, s, sum, a[1000002], ans;int main(){    scanf("%d", &t);    while(t--){        scanf("%d %d", &n, &s);        for(int i = 0; i < n; i++){            scanf("%d", &a[i]);        }        int t1 = 0, t2 = 0, sum = 0, ans = n + 1;        while(1){            while(sum < s && t2 < n)  sum += a[t2++];             if(sum < s)  break;            ans = min(ans, t2 - t1);            sum -= a[t1++];        }        if(ans == n + 1)  printf("0\n");        else  printf("%d\n", ans);    }    return 0;}

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