Max Sum of Max-K-sub-sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4836 Accepted Submission(s): 1765
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
題意:在組成環的n個數中求出不超過m個連續的數相加之和最大時的最大值、起始位置和終止位置(如果大於n則要減去n)
這個題我覺得還是有必要做一做的,雖然跟以前做的那個單調隊列題有點像,可是這個求和的我還是第一次做
#include<stdio.h> int a[100005],sum[200005],q[200005]; int main() { int t,n,m,i,j,maxx,e,s,first,last; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); sum[0]=0; for(i=1;i<=n;i++) { scanf("%d",&a[i]); sum[i]=sum[i-1]+a[i]; } for(;i<=n+m;i++) sum[i]=sum[i-1]+a[i-n];//向後延伸m,這樣就可以達到循環的效果 first=last=0; q[last++]=0;//這要注意,WA好多次了,還是對單調隊列理解的不夠 maxx=a[1]; e=s=1; for(i=1;i<=n+m;i++) { while(first<last&&sum[q[last-1]]>sum[i-1])//找最小值?因為我們看的是sum[i]-sum[q[first]],而sum[i]是定值 //所以要使這一項最大就必須使sum[q[first]]在m的范圍內最小 { last--; } q[last++]=i-1;//記錄的也是i-1 while(first<last&&q[first]<i-m)//超出范圍的去掉 first++; if(maxx<sum[i]-sum[q[first]]) { s=q[first]+1;//,我們當時記錄的是i-1,這裡要算起始點那就要加1 e=i;//終止點 maxx=sum[i]-sum[q[first]];//最大值 } } if(s>n)//這兩句話千萬別掉了 s-=n; if(e>n) e-=n; printf("%d %d %d\n",maxx,s,e); } return 0; }