Subsequence(尺取法)

Subsequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10890 Accepted: 4503

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

210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5

Sample Output

23

题目大意：起初的想法是先找到这个序列里最大的一个数m，然后在m前面的序列中找一个满足条件的最短子序列，然后在m后面的序列中找一个满足条件的子序列，比较这两者的长度，取小的，但这种做法是不对的，因为m成了分界，m在中间的子序列没被考虑在内，所以WA了好多次。。。汗！

正确的想法应该是，同时在m的前边找找，后面找找

附上错误的代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define MAX 100100int a[MAX],b[MAX];int main(){int t;scanf("%d",&t);while(t--){memset(a,0,sizeof(a));memset(b,0,sizeof(b));int n,s,i,m=-1,sum=0,la=100003,lb=100003;scanf("%d%d",&n,&s);for(i=0;i<n;i++){scanf("%d",&a[i]);sum+=a[i];if(a[i]>a[m]){m=i;}}    if(sum<s)    {    printf("0\n");    continue;    }          if(a[m]>=s)    {    printf("1\n");    continue;    }    int cnt=1;     //向前找    for(i=0;i<n;i++)      b[i]=a[i]; for(i=m;i>=0;i--){cnt++;a[i-1]+=a[i];if(a[i-1]>=s){la=cnt;break;}}  //向后找   cnt=1;  for(i=0;i<n;i++)        a[i]=b[i];  for(int j=m;j<n;j++) {   cnt++; a[j+1]+=a[j]; if(a[j+1]>=s) { lb=cnt; break; } } if(la!=100003||lb!=100003)  // if(la!=100003&&lb!=100003)第一次WA的地方   {    if(la<=lb)      printf("%d\n",la);    else      printf("%d\n",lb);  }  else    printf("0\n"); }return 0;}