POJ - 3061 Subsequence（連續子序列和>=s的最短子序列長度）

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

題目鏈接：http://poj.org/problem?id=3061

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題意：T組實例，每組實例給出n和s，接着n個數。求連續子序列和大於等於s的最短子序列長度。

分析：有點點模擬的意思，形象點說，就像你堆積木，超過一定值S后我們就把最下面的幾塊積木去掉。具體方法是從前面開始不斷累加，當和值超過S時減少前面的數值，然后記錄下剛好>=S的ans值，如此重復直到序列尾，輸出最小的ans。

 

ＡＣ代碼1：

 

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#include<time.h>
#include<stack>
using namespace std;
#define N 1200000
#define INF 0x3f3f3f3f

int dp[N];
int a[N];

int main()
{
    int n,m,T,i,j;

    scanf("%d", &T);

    while(T--)
    {
        scanf("%d %d", &n, &m);
        memset(dp,0,sizeof(dp));

        for(i=1;i<=n;i++)
        {
            scanf("%d", &a[i]);
            dp[i]=dp[i-1]+a[i];
        }

        int minn=INF,i=1;;
        for(j=1;j<=n;j++)
        {
             if(dp[j]-dp[i-1]<m)
                continue ;
             while(dp[j]-dp[i]>=m)
                i++;
             minn=min(minn,j-i+1);
        }

      if(minn==INF)
        printf("0\n");
      else
        printf("%d\n", minn);
    }
    return 0;
}

 

ＡＣ代碼2：（運用二分函數）

這里用到二分函數是找到剛好>=s的那個ans

 

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#include<time.h>
#include<stack>
using namespace std;
#define N 1200000
#define INF 0x3f3f3f3f

int dp[N];
int a[N];

int main()
{
    int n,m,T,i;

    scanf("%d", &T);

    while(T--)
    {
        scanf("%d %d", &n, &m);
        memset(dp,0,sizeof(dp));

        for(i=1;i<=n;i++)
        {
            scanf("%d", &a[i]);
            dp[i]=dp[i-1]+a[i];
        }

        int minn=INF;
        for(i=1;i<=n;i++)
            if(dp[n]-dp[i-1]>=m)
            {
                int ans=lower_bound(dp,dp+n,dp[i-1]+m)-dp;///二分函數 37                 minn=min(ans-i+1,minn);
            }

      if(minn==INF)
        printf("0\n");
      else
        printf("%d\n", minn);
    }
    return 0;
}