Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 105228 Accepted Submission(s): 24216
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
View Code
//第二组错误，输出了 7 7 #include <cstdio> #include <cstring> using namespace std; const int maxn = 100010; int ans[maxn] = {0}; int b[maxn][2] = {0}; int a[maxn]; int main() { int i,j,k; int T; scanf("%d",&T); int n,m; int num; int flag = 1; int temp = T; while(T--) { scanf("%d",&num); memset(ans,0,sizeof(ans)); memset(b,0,sizeof(b)); memset(a,0,sizeof(a)); for(i=1; i<=num; i++) scanf("%d",a+i); ans[1] = a[1]; b[1][0] = b[1][1] = 1; int max = -1; int next; for(i=2; i<=num; i++) { if(ans[i-1]>0) { ans[i] = ans[i-1] + a[i]; //if(a[i]>0) { b[i][0] = b[i-1][0]; b[i][1] = i; } // else // { // b[i][0] = b[i-1][0]; // b[i][1] = i-1; // } } else { ans[i] = a[i]; b[i][0] = b[i][1] = i; } if(ans[i]>max) { max = ans[i]; next = i; } } printf("Case %d:\n",flag); printf("%d %d %d\n",max,b[next][0],b[next][1]); if(flag<temp) printf("\n"); flag++; } return 0; }
1 #include <cstdio> 2 #include <cstring> 3 using namespace std; 4 5 const int maxn = 100010; 6 int a[maxn] = {0}; 7 8 int main() 9 { 10 int i,j,k; 11 int T; 12 scanf("%d",&T); 13 int num; 14 int flag = 1; 15 int sum = 0; 16 int start,end,last; 17 int temp = T; 18 while(T--) 19 { 20 21 int max = (1<<31); 22 scanf("%d",&num); 23 memset(a,0,sizeof(a)); 24 sum = 0; 25 last = 1; 26 for(i=1; i<=num; i++) 27 { 28 scanf("%d",a+i); 29 sum += a[i]; 30 if(sum>max) 31 { 32 start = last; 33 max= sum; 34 end = i; 35 } 36 if(sum<0) 37 { 38 sum = 0; 39 last = i+1; 40 } 41 42 } 43 printf("Case %d:\n",flag); 44 printf("%d %d %d\n",max,start,end); 45 if(flag<temp) 46 printf("\n"); 47 flag++; 48 } 49 while(1); 50 return 0; 51 }
