Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6844 Accepted Submission(s): 2518
Problem Description
Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1]. Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.
Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the minimum length of them.
Sample Input
4 6 3 6 -1 2 -6 5 -5 6 4 6 -1 2 -6 5 -5 6 3 -1 2 -6 5 -5 6 6 6 -1 -1 -1 -1 -1 -1
Sample Output
7 1 3 7 1 3 7 6 2 -1 1 1
题解:
让找长度小于等于k的最长字串合;
暴力超时;
超时代码:
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef long long LL; const int MAXN=100010; int m[MAXN]; int main(){ int T,N,K; SI(T); while(T--){ SI(N);SI(K); for(int i=1;i<=N;i++)SI(m[i]); int cur=0,ans=-INF,l,r; for(int i=1;i+K-1<=N;i++){ cur=0; for(int j=0;j<K;j++){ cur+=m[i+j]; if(cur>ans){ ans=cur; l=i;r=i+j; } } } printf("%d %d %d\n",ans,l,r); } return 0; }