poj--3061--Subsequence(贪心)
2016-03-05 14:04
387 查看
Subsequence
Description
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.
Input
The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The
input will finish with the end of file.
Output
For each the case the program has to print the result on separate line of the output file.if no answer, print 0.
Sample Input
Sample Output
Source
Southeastern Europe 2006
简直就是超时专题,还是自己太傻了,想着暴力可以的,
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 10882 | Accepted: 4498 |
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.
Input
The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The
input will finish with the end of file.
Output
For each the case the program has to print the result on separate line of the output file.if no answer, print 0.
Sample Input
2 10 15 5 1 3 5 10 7 4 9 2 8 5 11 1 2 3 4 5
Sample Output
2 3
Source
Southeastern Europe 2006
简直就是超时专题,还是自己太傻了,想着暴力可以的,
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int num[100000+10]; int main() { int t; scanf("%d",&t); while(t--) { int n,ans; scanf("%d%d",&n,&ans); memset(num,0,sizeof(num)); for(int i=0;i<n;i++) scanf("%d",&num[i]); int l=0,r=0,minn=0x3f3f3f,sum=0; bool f=false; while(r<n) { sum+=num[r++]; while(sum>=ans) { f=true; minn=min(minn,r-l); sum-=num[l++]; } } if(f) printf("%d\n",minn); else printf("0\n"); } return 0; }
相关文章推荐
- UESTC 758-P酱的冒险旅途【BFS】
- 今日开讲—— easyui-combobox动态赋值
- LeetCode 51 - N-Queens II
- Ngui 拖拽到底部,直接循环到顶部
- Uva1594 Ducci Sequence
- poj Subsequence 3061 (高效&DP)
- LeetCode 51 - N-Queens
- Android几种在其他线程中更新UI的方法
- UUID 和 GUID 的区别
- UESTC P酱的冒险旅途 785 (规律模拟)
- 利用系统自带的UITabBarController纯代码搭建TabBar
- UESTC 1251-谕神的密码【模拟】
- 内核工作队列workqueue
- UESTC--758--P酱的冒险旅途(模拟)
- Subsequence(暴力+二分)
- 优先队列:priority_queue
- 利用Runtime自定义TextField
- String/StringBuffer/StringBuilder的区别
- druid配置
- easy-UI作为页面展示的一个例子