HDU 1003 Max Sum （经典DP）

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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

前几天一直在刷图论的题，有点疲倦，做做有意思的题目涨涨姿势

第一次手敲DP，下面附上代码

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int dp[111111],a[111111];int main(){int t,n,i,j;scanf("%d",&t);for(int z=0;z<t;z++) {if(z) printf("\n");scanf("%d",&n);for(i=1;i<=n;i++) scanf("%d",&a[i]);dp[1]=a[1];for(i=2;i<=n;i++) dp[i]=max(dp[i-1],0)+a[i];  //最大连续字串和  int mmax=-99999999,index=-1;for(i=1;i<=n;i++) {if(dp[i]>mmax) {mmax=dp[i];index=i;}}for(i=index;i>0;i--) {if(dp[i]<0) break;}if(mmax<0) i=index-1;printf("Case %d:\n",z+1);printf("%d %d %d\n",mmax,i+1,index);}return 0;}

下面再附上别人的代码，节省了很多空间，方法是用模拟做

#include <stdio.h>int main(){int n,T,a,sta,end,max,k,i,p,t;  scanf("%d",&T); for(p=1;p<=T;p++) {  scanf("%d",&n);  max=-9999;     //因为一个数a 是-1000~1000的，所以这里相当于变成最小值  t=0;           //表示 某段连续和  sta=end=k=1;   // sta最大和的开始,end最大和的结束,k记录每次求和的开始  for(i=1;i<=n;i++) {   scanf("%d",&a);        t+=a;           if(t>max) {   //记录最大连续和的值    max=t;    sta=k;    end=i;   }   if(t<0) {      t=0;    k=i+1;   }  }    if(p!=1) printf("/n");  printf("Case %d:/n",p);  printf("%d %d %d/n",max,sta,end); }}

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