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；
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <time.h> 5 #include <math.h> 6 #include <queue> 7 #include <stack> 8 #include <list> 9 #include <set> 10 #include <map> 11 #include <vector> 12 #include <algorithm> 13 #include <iostream> 14 using namespace std; 15 16 const int NO = 100000+10; 17 int sum[NO],a[NO]; 18 int n, num; 19 20 void solve() 21 { 22 for( int i = 0; i < n; ++i ) 23 sum[i+1] = sum[i] + a[i]; 24 if( sum[n] < num ) { 25 puts( "0" ); 26 return; 27 } 28 int ret = n; 29 for( int x = 0; sum[x] + num <= sum[n]; ++x ) 30 { 31 int t = lower_bound( sum+x, sum+n, sum[x]+num ) - sum; 32 ret = min( ret, t-x ); 33 } 34 printf( "%d\n", ret ); 35 } 36 37 int main() 38 { 39 int T; 40 scanf( "%d", &T ); 41 while( T-- ) 42 { 43 memset( sum, 0, sizeof( sum ) ); 44 scanf( "%d%d", &n, &num ); 45 for( int i = 0; i < n; ++i ) 46 scanf( "%d", &a[i] ); 47 solve(); 48 } 49 return 0; 50 }