POJ3061-Subsequence

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首先是一个nlogn的解法：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100000+5;int num[maxn];int sum[maxn];int main(int argc, char const *argv[]) {    int t;    scanf("%d", &t);    while (t--) {        int n, s;        scanf("%d%d", &n, &s);        for (int i = 0; i < n; i++) {            scanf("%d", &num[i]);            sum[i+1] = sum[i] + num[i];        }        if (sum[n] < s) {            printf("0\n");            continue;        }        int res = n;        for (int i = 0; sum[i] + s <= sum[n]; i++) {            int t = lower_bound(sum + i, sum + n, sum[i] + s) - sum;            res = min(res, t - i);        }        printf("%d\n", res);    }    return 0;}

另外一种更高效的算法是尺取法，效率为O(n)，与上面类似的是斗士在区间上反复对比考量。仔细阅读其实并不难理解。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100000+5;int a[maxn];int main(int argc, char const *argv[]) {    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 res = n + 1;        int j = 0, k = 0, sum = 0;        while (true) {            while (k < n && sum < s) {                sum += a[k++];            }            if (sum < s)                break;            res = min(res, k - j);            sum -= a[j++];        }        if (res > n)            res = 0;        printf("%d\n", res);    }    return 0;}

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