Subsequence

SubsequenceTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 16645Accepted: 7068DescriptionA 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.InputThe 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.OutputFor each the case the program has to print the result on separate line of the output file.if no answer, print 0.Sample Input210 155 1 3 5 10 7 4 9 2 85 111 2 3 4 5Sample Output23

个人理解：

尺取法。只要sum<s就一直加，直到sum>=s，去掉前面的项，直到找到最小项

代码：

#include<iostream>#include<algorithm>using namespace std;#define max  100005int N,S,T;int a[max];void input(){  cin>>N>>S;  for(int i=0;i<N;i++)      cin>>a[i];}//尺取法void solve(){    int sum=0,Start=0,End=0,ans=max;    while(true){        while(End<N&&sum<S){ //当总和sum<S时，一直往后加            sum+=a[End];            End++;        }        if(sum<S) break;        while(Start<N&&sum>=S){ //当总和sum>=S时，一直去掉前面的项            sum-=a[Start];            Start++;        }        ans=min(ans,End-Start+1);    }    if(ans>N) cout<<"0"<<endl;    else cout<<ans<<endl;}int main(){    cin>>T;    while(T--){        input();        solve();    }return 0;}