UVALive - 2678 Subsequence

最新推荐文章于 2019-09-06 15:23:09 发布

原创最新推荐文章于 2019-09-06 15:23:09 发布 · 381 阅读

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本文介绍了一种寻找连续子序列使和大于等于给定值的算法问题，提供了两种解决方案，一种为nlog(n)复杂度，另一种为O(n)复杂度。

Subsequence

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

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

Many test cases will be given. 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 there isn't such a subsequence, print 0 on a line by itself.

Sample Input

10 15 
5 1 3 5 10 7 4 9 2 8 
5 11 
1 2 3 4 5

Sample Output

2 
3

Source

Regionals 2006 >> Europe - Southeastern

分析：

解法一：

ac代码：nlog（n）

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100000+10;
int a[maxn],b[maxn];
int n,s;
int main()
{
while(scanf("%d%d",&n,&s)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
b[0]=0;
for(int i=1;i<=n;i++)
b[i]=b[i-1]+a[i];
int ans=n+1;
for(int j=1;j<=n;j++)
{
int i=lower_bound(b,b+j,b[j]-s)-b;
if(i>0) ans=min(ans,j-i+1);
}
printf("%d\n",ans==n+1?0:ans);
}
return 0;
}

解法二：O(n)

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100000+10;
int a[maxn],b[maxn];
int n,s;
int main()
{
while(scanf("%d%d",&n,&s)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
b[0]=0;
for(int i=1;i<=n;i++)
b[i]=b[i-1]+a[i];
int ans=n+1;
int i=1;
for(int j=1;j<=n;j++)
{
if(b[i-1]>b[j]-s) continue;
while(b[i]<=b[j]-s) i++;
ans=min(ans,j-i+1);
}
printf("%d\n",ans==n+1?0:ans);
}
return 0;
}