Problem Description
Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second 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 first one. Output a blank line between two cases.
Sample Input
2 5 6 -1 5 4 -7 7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4 Case 2: 7 1 6
超时代码
1 #include<stdio.h> 2 int main() 3 { 4 int a[100000],T,N,T1,j,i; 5 scanf("%d",&T); 6 T1=T; 7 while(T--) 8 { 9 int Msum=0,sum=0,s=0,w=0; 10 printf("case %d:\n",T1-T); 11 scanf("%d",&N); 12 for(i=0;i<N;i++) 13 scanf("%d",&a[i]); 14 Msum=a[0]; 15 for(j=0;j<N;j++) 16 { 17 for(i=j;i<N;i++) 18 { 19 if(a[i]<=0) 20 { 21 sum+=a[i]; 22 continue; 23 } 24 sum+=a[i]; 25 if(Msum<sum) 26 { 27 s=j; 28 w=i; 29 Msum=sum; 30 } 31 } 32 sum=0; 33 } 34 printf("%d %d %d\n\n",Msum,s+1,w+1); 35 } 36 return 0; 37 }
AC代码
1 /*状态转移方程 d[i] = max(d[i-1]+a[i], a[i]) 2 d[i]表示以i位置结束的最大子序列之和。*/ 3 #include<stdio.h> 4 int main() 5 { 6 int a[100000]; 7 int T,T1; 8 scanf("%d",&T); 9 T1=T; 10 while(T--) 11 { 12 int sum=0,msum=0,i,x=0,y=0,start=0,end=0,N; 13 scanf("%d",&N); 14 for(i=0;i<N;i++) 15 scanf("%d",&a[i]); 16 sum=a[0]; 17 msum=sum; 18 for(i=1;i<N;i++) 19 { 20 if(sum<0)/*dp[i-1]对a[i]不仅没有贡献,反而有损害,就应该舍弃*/ 21 { 22 x=y=i; 23 sum=a[i]; 24 } 25 else 26 { 27 sum+=a[i]; 28 y=i; 29 } 30 if(sum>msum) 31 { 32 msum=sum; 33 start=x; 34 end=y; 35 } 36 } 37 printf("Case %d:\n",T1-T); 38 39 if(T==0) 40 printf("%d %d %d\n",msum,start+1,end+1); 41 else 42 printf("%d %d %d\n\n",msum,start+1,end+1); 43 44 } 45 }