Subsequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 24179 Accepted: 10231 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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <utility> 6 #include <stack> 7 #include <map> 8 #include <queue> 9 #include <vector> 10 #include <cmath> 11 #include <iostream> 12 #include <algorithm> 13 #define pi acos(-1.0) 14 #define e 2.718281828 15 #define lowbit(x) (x&(-x)) 16 #define ll long long 17 const int N =1e5+1000; 18 const ll inf = 0x3f3f3f3f; 19 using namespace std; 20 int t; 21 ll n,s; 22 ll a[N]; 23 int main() 24 { 25 scanf("%d",&t); 26 while(t--) 27 { 28 scanf("%lld%lld",&n,&s); 29 for(ll i=0;i<n;i++) scanf("%lld",&a[i]) ; 30 ll st=0,en=0,sum=0; 31 ll ans = inf; 32 while(1) 33 { 34 while(en<n&&sum<s){ 35 sum+=a[en++]; 36 } 37 if(sum<s) break; 38 ans=min(ans,en-st);//en一定>=st 39 sum-=a[st++]; 40 } 41 if(ans==inf) ans=0; 42 printf("%lld ",ans); 43 } 44 return 0; 45 }