Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 237978 Accepted Submission(s): 56166

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

Author

Ignatius.L

Recommend

这道题目写的时候WA了很多次

最开始用结构体写的，写了两层循环，然后自己琢磨的瞎改，一直都是两层循环

#include
      
       #include
       
        #include
        
         #include
         
          #include
          #include
           
            using namespace std;int a[100010];struct node{ int num1,num2,maxn;} b[100010];bool cmp(struct node a,struct node b){ return a.maxn > b.maxn;}int main(){ int t; scanf("%d",&t); for(int p=1; p<=t; p++) { int n; scanf("%d",&n); for(int i=1; i<=n; i++) { b[i].num1=0; b[i].num2=0; b[i].maxn=0; } for(int i=1; i<=n; i++) scanf("%d",&a[i]); int j=1; for(int k=1; k<=n; k++) { if(a[k]>=0) { if(j!=1) j++; b[j].maxn += a[k]; b[j].num1 = k; int f = k; for(int i=k+1; i<=n; i++) { if(a[i]<0) { b[j].num2 = i-1; int d = j; b[++j].maxn = b[d].maxn + a[i]; b[j].num1 = f; } if(a[i]>=0) { b[j].maxn += a[i]; } if(i==n) { if(a[i]>=0) { b[j].maxn += a[i]; b[j].num2 = i; } } } } } /*for(int i=1;i

后来我师傅指导了我，直接用一层循环就解决了。。

就是用sum不停累加，如果sum小于0了，就重置sum等于当前的a[i]，同时更新起始，结束位置

（注意sum最大时有可能为负数，所以定义maxn时要注意maxn取值要为最小--1001 ）

#include
      
       #include
       
        #include
        
         #include
         
          #include
          #include
           
            using namespace std;int a[100010];int main(){ int t; scanf("%d",&t); for(int p=1; p<=t; p++) { int n; scanf("%d",&n); for(int i=1; i<=n; i++) scanf("%d",&a[i]); int sum=0,num1=1,tmp=1,num2=1,maxn=-10000000;//注意maxn的取值 for( int i = 1 ; i <= n ; i++ ) { sum += a[i] ; if( sum > maxn )//同时跟新最大值，起始位置，结束位置 { maxn = sum ; num2 = i ; num1 = tmp ; } if( sum < 0 ) { sum = 0 ; tmp = i + 1 ; } } printf("Case %d:\n",p); printf("%d %d %d\n",maxn,num1,num2); if(p!=t) printf("\n");//还有这个输出要注意！！！最后一组数据不要输出多余空行，其他还要多输出一行空行 } return 0;}

这段时间学dp学到的dp解法

#include 
      
       #include
       
        #include
        
         #include
         
          using namespace std;int a[100010],sum[100010],s[100010];int main(){    int T;    cin >> T;    for(int k=1;k<=T;k++){        int n;        cin >> n;        for(int i=0;i
          
           > a[i];        }        int ans = 0;        sum[0] = a[0];        s[0] = 0;        for(int i=0;i
           
            = 0){ //只要不是小于零就可以继续加，记下每次加后得到的值 sum[i] = sum[i-1] + a[i]; s[i] = s[i-1]; } else{ //小于零重新开始累加 sum[i] = a[i]; s[i] = i; } if(sum[ans] < sum[i]){ //求出记录中的最大值 ans = i; } } cout << "Case " << k << ":" << endl; cout << sum[ans] << " " << s[ans]+1 << " " << ans + 1 << endl; if(k!=T){ cout << endl; } } return 0;}

转载于:https://www.cnblogs.com/l609929321/p/6596196.html

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