POJ 3061 Subsequence
来源:互联网 发布:网络买古瓷咋卖 编辑:程序博客网 时间:2024/04/27 16:59
Subsequence
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15819 Accepted: 6675
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
210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5
Sample Output
23
题目链接:http://poj.org/problem?id=3061
题意:给定长度为n的整数数列以及整数S,求总和不小于S的连续子序列长度的最小值。如果杰不存在,则输出0。
解题思路:采用尺取法,设以a[s]开始的最初大于S的连续子序列为a[s]+a[s+1]+...+a[t-1],这时a[s+1]+a[s+2]+...+a[t-2]<S,所以从a[s+1]开始总和最初超过S的连续子序列如果是a[s+1]+a[s+2]+...+a[t‘-1]的话,则必然有t<=t'。利用这一性质便可以设计出这样的算法:1、以s=t=sum=0初始化。2、只要依然有sum<S,就不断将sum增加a[t],并将t增1。3、如果2中无法满足sum>=S的条件的话就跳出循环。否则更新ans=min(ans,t-s)。4、将sum减去a[s],s增加1后回到2。(可以参考挑战程序设计竞赛)
AC代码:
#include<cstdio>#include<algorithm>using namespace std;const int MAXN = 1e5+7;int n,S;int a[MAXN];void solve(){int ans=n+1;int sum=0,s=0,t=0;for(;;){while(t<n&&sum<S){sum+=a[t++];}if(sum<S) break;ans=min(ans,t-s);sum-=a[s++];}if(ans>n) ans=0;printf("%d\n",ans);}int main(void){int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&S);for(int i=0;i<n;i++){scanf("%d",&a[i]);}solve();}return 0;}
阅读全文
0 0
- poj 3061 Subsequence
- POJ-3061-Subsequence
- POJ - 3061 Subsequence
- POJ:3061 Subsequence
- poj 3061 Subsequence
- poj 3061 Subsequence
- POJ 3061 Subsequence
- poj 3061 Subsequence
- poj 3061 Subsequence
- POJ 3061 Subsequence
- POJ 3061 Subsequence
- poj 3061 Subsequence
- poj 3061 Subsequence
- poj 3061 Subsequence
- POJ-3061 Subsequence
- poj 3061 Subsequence
- POJ 3061 Subsequence
- POJ 3061 Subsequence《《前加后去,,,》》
- FZU 1683 纪念SlingShot
- [普通平衡树]多种姿势 栽培小树苗
- Gym
- 数组的使用以及常见的方法
- 选择问题
- POJ 3061 Subsequence
- Hbase学习以及hbase下表的增删查操作
- HDU5631 Rikka with Graph
- mySQL查询数据库的内容
- flask数据库查询且排序
- 微服务(Microservice)追记
- 1.25Android 学习+进度之二十五-小修小补及完工
- perl 退出函数问题
- 数值处理