首页 > 代码库 > poj3061 Subsequence ,尺取法
poj3061 Subsequence ,尺取法
A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.
尺取法
设置两个指针s,t。
维护一个区间的信息,然后根据指针的移动,更新区间信息。
时间复杂度 O(n).
同类型练习题:
poj 3320 Jessica‘s Reading Problem
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn = 100000 + 100; int a[maxn]; int main() { int T, n, S; scanf("%d", &T); while(T--) { scanf("%d%d", &n, &S); for(int i=1; i<=n; ++i) scanf("%d", &a[i]); int ans = n+1; int tmp = 0, s = 1, t = 1; while(true) { while(t<=n && tmp < S) { tmp += a[t]; t++; } if(tmp < S) break; ans = min(ans, t-s); tmp -= a[s++]; } if(ans<=n) printf("%d\n", ans); else printf("0\n"); } return 0; }
poj3061 Subsequence ,尺取法
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。