题目链接：uva 1121 - Subsequence
题目大意：给出一个序列，要求求出序列中最短子序列的元素和不小于k的长度。
解题思路：r维护加入，l维护减除，r - l即为当前个数。
#include#include #include using namespace std; const int N = 100005; int n, k, num[N]; void init() { for (int i = 0; i < n; i++) scanf("%d", &num[i]); } int solve() { int ans = n + 1, sum = 0; int l = 0, r = 0; while (r < n) { sum += num[r++]; while (sum >= k) { ans = min(ans, r - l); sum -= num[l++]; } } return ans == n + 1 0 : ans; } int main() { while (scanf("%d%d", &n, &k) == 2) { init(); printf("%d\n", solve()); } return 0; }