POJ 3061

题意：给你 n 个正整数以及整数s，求出总和不小于 s 的连续子序列的长度最小值。若无则输出 0 ；

题解：枚举 + 二分；

 

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;

const int NO = 100000+10;
int sum[NO],a[NO];
int n, num;

void solve()
{
    for( int i = 0; i < n; ++i )
        sum[i+1] = sum[i] + a[i];
    if( sum[n] < num ) {
        puts( "0" );
        return;
    }
    int ret = n;
    for( int x = 0; sum[x] + num <= sum[n]; ++x )
    {
        int t = lower_bound( sum+x, sum+n, sum[x]+num ) - sum;
        ret = min( ret, t-x );
    }
    printf( "%d\n", ret );
}

int main()
{
    int T;
    scanf( "%d", &T );
    while( T-- )
    {
        memset( sum, 0, sizeof( sum ) );
        scanf( "%d%d", &n, &num );
        for( int i = 0; i < n; ++i )
            scanf( "%d", &a[i] );
        solve();
    }
    return 0;
}

 

法二：尺取法：反复推进区间的开头和结尾，来求取满足条件的最小区间。

 

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <vector>
#include <list>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;

const int NO = 100000 + 10;
int a[NO];
int n, s;

void solve()
{
    int sum = 0, t = 0, x = 0;
    int res = n + 1;
    while( true )
    {
        while( t < n && sum < s )
            sum += a[t++];
        if( sum < s ) break;
        res = min( t-x, res );
        sum = sum - a[x++];
    }
    if( res > n ) res = 0;
    printf( "%d\n", res );
}

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;
}