POJ 3061

Subsequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7854		Accepted: 3021

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

Source

Southeastern Europe 2006

 

两种方法

一种是用sum[i]数组存储从1 到 i 的和，然后二分长度。

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

#define maxn 100005

int n;
int a[maxn],s,sum[maxn];

bool check(int x) {
        for(int i = 1; i + x - 1 <= n; ++i) {
                if(sum[i + x - 1] - sum[i - 1] >= s) return true;
        }

        return false;
}



void solve() {
        int l = 1,r = n + 1;

        while(l < r) {
                int mid = (l + r) >> 1;
                if(check(mid)) r = mid;
                else l = mid + 1;
        }

        printf("%d
",l == n + 1 ? 0 : l);
}

int main()
{
    //freopen("sw.in","r",stdin);

    int t;
    scanf("%d",&t);

    while(t-- ) {
            memset(sum,0,sizeof(sum));
            scanf("%d%d",&n,&s);
            for(int i = 1; i <= n; ++i) {
                    scanf("%d",&a[i]);
                    sum[i] += sum[i - 1] + a[i];
            }

            solve();
    }

    return 0;
}

第二种是假设从s位置开始到t的和大于S，并且s + 1 到t' 的和大于S，则t‘ > t 由此可以跑一次o(n)解决

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

#define maxn 100005

int n,S;
int a[maxn];

void solve() {
        int s = 1,sum = 0,pos = 1;
        int ans = n + 1;

        for(; ; ++s) {
            while(sum <= S && pos <= n) {
                sum += a[pos++];
            }
            //printf("p = %d s = %d
",pos);
            if(sum < S) break;
            sum -= a[s];
            ans = min(ans,pos - s);
        }

        printf("%d
",ans == n + 1 ? 0 : ans);

}

int main() {
        //freopen("sw.in","r",stdin);

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

                solve();
        }

        return 0;
}