POJ——3061Subsequence（尺取法或二分查找）

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11224		Accepted: 4660

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

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

Sample Output

2
3

尺取法做法：一直向前增加num[r]，直到不能增加，再判断r-l，然后再一直向前减去num[l]，然后判断是否小于s。一直循环。70+ms

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int N=100010;
int list[N];
inline int Scan()     
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')
        flag=1;
    else if(ch>='0'&&ch<='9')
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}
int main (void)
{
	int t,n,s,i,j,temp,sum,ans;
	scanf("%d",&t);
	while (t--)
	{
		memset(list,0,sizeof(list));
		scanf("%d%d",&n,&s);
		sum=0;
		for (i=1; i<=n; i++)
		{
			scanf("%d",&list[i]);
		}
		int l,r,dx;
		l=r=1;
		temp=0;
		dx=N;
		while (1)
		{
			while (r<=n&&temp<s)//r向前递增
			{
				temp+=list[r++];
			}
			if(temp<s)
				break;
			dx=min(r-l,dx);
			temp-=list[l++];//l向前递增
		}
		if(dx==N)//特判
			puts("0");
		else
			printf("%d
",dx);
	}
	return 0;
}

二分查找做法：建立另外一个数组sufix来储存前缀和，然后题目变成了保证sufix[r]-sufix[l-1]>=S这样的条件下求r-l的最小值，显然r>=l。将这个式子移项得到sufix[r]>=S+sufix[l]。

 就是说当遍历l的时候要使r最小，然后就用自带的lowerbound函数来求。

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int N=100010;
int list[N];
int sufix[N];
int main (void)
{
	ios::sync_with_stdio(false);
	int t,n,s,i,j,temp,sum,ans;
	cin>>t;
	while (t--)
	{
		cin>>n>>s;
		for (i=1; i<=n; i++)
		{
			cin>>list[i];
			sufix[i]=sufix[i-1]+list[i];
		}
		int dx=N;
		if(sufix[n]<s)//特判
		{
			cout<<0<<endl;
			continue;
		}
		for (i=0; sufix[i]+s<=sufix[n]; i++)//由于是公式内为l-1，因此i要从0开始，即
		{
			int t=lower_bound(sufix+i+1,sufix+n+1,s+sufix[i])-(sufix+i);
			dx=min(t,dx);
		}
		cout<<dx<<endl;
	}
	return 0;
}