POJ 3061-Subsequence(尺取法,二分)

Subsequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15518 Accepted: 6554

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

//二分思想//先把a1+a2+...+ai的和存到sum[i]//循环i,i从1开始,用二分法找到以第i个元素开始,至少要加几个元素能大于s#include <stdio.h>#include <string.h>#define maxn 100005int a[maxn];int sum[maxn];int main(){int T;scanf("%d", &T);while(T--){int n, s;memset(sum, 0, sizeof(sum));scanf("%d%d", &n, &s);for(int i=1; i<=n; i++){scanf("%d", &a[i]);sum[i] = a[i]+sum[i-1];}int ans = 0x3f3f3f3f;for(int i=1; i<=n; i++){int l=i, r=n, mid;int ok = 0;while(r>=l){mid = (r+l)>>1;if((sum[mid] - sum[i-1])>=s){    ok = 1;    r = mid - 1;}elsel = mid + 1;}if(r-i+2<ans && ok)ans = r-i+2;           // printf("i==%d  r=%d, l=%d. mid=%d, ll = %d,ok = %d\n", i,r,l,mid,r-i+2, ok);}printf("%d\n", ans==0x3f3f3f3f?0:ans);}return 0;}