【二分答案nlogn/标解O(n)】【UVA1121】Subsequence

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

有n个正整数组成一个序列。给定整数S，求长度最短的连续序列，使它们的和大于或等于S。

【输入格式】

输入包含多组数据。每组数据的第一行为整数n和S（10＜n≤100 000，S＜10⁹）；第二行为n个正整数，均不超过10 000。输入结束标志为文件结束符（EOF）。

【输出格式】

对于每组数据，输出满足条件的最短序列的长度。如果不存在，输出0。

二分答案(即序列长度)+前缀和可以轻松搞定

复杂度为O（n*logn）

（这个方法可以解决数可能为负数的数据，所以也是一种值得记忆的方法）

#include <cstdio>  
#include <cstdlib>  
#include <cmath>  
#include <cstring>  
#include <ctime>  
#include <algorithm>  
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313   
using namespace std;
int n,S;
int A[100010];
int ans;
int OK(int m)
{
	if(S==0) return 1;
	int temp=0;
	for(int i=1;i<=n-(m-1);i++)
	{
		temp=A[i+(m-1)]-A[i-1];
		if(S<=temp) return 1;
	}
	return 0;
}
void solve()
{
	ans=0;
	int l=1,r=n;
	int m=(l+r)/2;
	while(l<r)
	{		
		if(OK(m)) r=m;
		else l=m+1;
		m=(l+r)/2;
	}
	ans=m;
}
void input()
{
	for(int i=1;i<=n;i++)
		{
			scanf("%d",&A[i]);
			A[i]=A[i-1]+A[i];
		}
}
int main()
{
	while(scanf("%d%d",&n,&S)!=EOF)
	{
		input();
		solve();
		if(ans!=n)
		printf("%d\n",ans);
		else if(A[n]<S)
		{
			printf("0\n");
		}
		else printf("%d\n",n);
	}
	return 0;
}