hdu1003

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98850 Accepted Submission(s): 22737

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

贪心算法：

#include<stdio.h>int main(){    int m; //case数    int n; // 数列长度    int i,j;    int p,start,temp,end;    int max,sum;    scanf("%d",&m);    for(i = 1; i <= m; i++)    {        sum = 0;        temp = 1;            // 扫描到的第一个子段肯定是起始位置为1        max = -1001;        scanf("%d",&n);        for(j = 1; j <= n; j++)        {            scanf("%d",&p);            sum += p;            if(sum > max)            {                max = sum;      // 遇到更大的子段和，更新起始，结束位置                start = temp;                end = j;            }            if(sum < 0)            {                sum = 0;       //当前子段和归零                temp = j + 1;   //下一位置为起始位            }        }        printf("Case %d:\n%d %d %d\n",i,max,start,end);        if(i < m) printf("\n");    }    return 0;}

动态规划

#include<cstdio>using namespace std;int dp[100010]; // dp[i]存储以第i个元素为结尾的最大子段和int main(){        int T;    scanf("%d", &T);    for(int I = 1;I <= T; I++)    {        int n,start = 1,end = 1,ans, temp = 1;        scanf("%d",&n);        for(int i = 1;i <= n;i ++)            scanf("%d",dp + i);

        ans = dp[1];        for(int i = 2;i <= n;i ++)        {            int t = dp[i] + dp[i - 1];            if(t >= dp[i])//如果加上第i个元素比之前的和大，

                dp[i] = t;//那么第i个元素为结尾的最大子段和            else//就是第i个元素和前一个元素为结尾的最大子段和的和             temp = i;            if(ans < dp[i])//如果以第i个元素为结尾的最大子段和比当前最大的还大，就更新            {                ans = dp[i];                start = temp;                end = i;            }        }        printf("Case %d:\n",I);        printf("%d %d %d\n",ans,start,end);        if(I < T)printf("\n");    }    return 0;}

通过滚动数组减少内存的动态规划

#include<cstdio>using namespace std;int main(){     int T, dp[2];//因为再上一个代码里面dp[i]只是和前一个元素dp[i - 1]有关系，    scanf("%d", &T);//所以没必要开那么大的数组，每次用完i-1个元素就可以覆盖它    for(int I = 1;I <= T; I++)    {        int n, start = 1, end = 1, ans, temp = 1;        scanf("%d",&n);        scanf("%d",&ans);        dp[0] = ans;        for(int i = 2,j = 1;i <= n;i ++)        {            scanf("%d",dp + j);            int t = dp[j] + dp[j ^ 1];            if(t >= dp[j])                dp[j] = t;            else             temp = i;            if(ans < dp[j])            {                ans = dp[j];                start = temp;                end = i;            }           j ^= 1;        }        printf("Case %d:\n",I);        printf("%d %d %d\n",ans,start,end);        if(I < T)printf("\n");    }    return 0;}