poj 3061 Subsequence 【尺取法 or STL lower_bound】


Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9729		Accepted: 3914

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-->挑战详解

我们设以a[s] 开始总和最初大于S时的连续子序列为a[s] + ... + a[t-1],这时a[s+1] + ... + a[t-2] < a[s] + ... + 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[t'-1]的话，则必然有t <= t'。

算法思路：

（1）初始化s = t = sum = 0.

（2）只要依然有sum < S，就不断将sum增加a[t]，并将t增加一。

（3）如果（2）中无法满足sum >= S则终止。否则 更新结果ans = min(ans, t-s)。

（4）将sum减去a[t]，s增加1然后回到（2）。

代码实现:时间复杂度(n)

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[100000+10];
int n, S;
void solve()
{
	int t, sum, s;//s代表起始位置 
	int ans = n + 1;//最后结果 
	t = sum = s = 0;
	for(;;)
	{
		while(t < n && sum < S)//知道sum >= S 或者 t >= n为止 
		{
			sum += a[t++];//t为序列末位置 
		}
		if(sum < S) 
		break;//s为起始位置的情况下找不到连续的数之和大于或者等于S 
		ans = min(ans, t-s); 
		sum -= a[s++];//去掉a[s]，起始位置s往后走 
	}
	if(ans > n)
	ans = 0;
	printf("%d\n", ans);
}
int main()
{
	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;
}

STL lower_bound: 时间复杂度（nlog(n)） 请先会用lower_bound

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX 100000+10
using namespace std;
int sum[MAX];
int main()
{
    int t, n, S, i, j;
    int Min;//最小长度 
    int first, pos;//序列初位置 末位置 
    int a;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &S);
        sum[0] = 0;
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &a);
            sum[i] = sum[i-1] + a;
        }
        if(sum
 < S)
        {
            printf("0\n");
            continue;
        }
        Min = n+1;
        for(first = 0; sum[first] + S <= sum
; first++)
        {
            //upper_bound  
            pos = lower_bound(sum+first, sum+n, sum[first]+S) - sum; 
            Min = min(Min, pos-first); 
        }
        printf("%d\n", Min);
    }
    return 0;
}