hdu 1003 Max Sum(简单dp)

题目大意是，给你一个数字序列，求其子串的最大和，注意是子串，子串是连续的字符

注意题目要求：If there are more than one result, output the first one. Output a blank line between two cases.

动规方程 dp[i]=(dp[i-1]+a[i]>a[i])?dp[i-1]+a[i]:a[i];
其实就等价于(dp[i-1]>0)?dp[i-1]+a[i]:a[i];
dp[i]即前i个数字的子串最大和
上代码

#include <iostream>#include <string.h>using namespace std;int dp[100005],a[100005];int main() {    int t,count=0;    cin>>t;    for(count=1; count<=t; count++) {        memset(dp,0,sizeof(dp));        memset(a,0,sizeof(a));        int i,n,max,start,end;        cin>>n;        for(i=1; i<=n; i++)            cin>>a[i];        dp[1]=max=a[1];        for(i=2; i<=n; i++) {            dp[i]=dp[i-1]>0?dp[i-1]+a[i]:a[i];        }        end=1;        for(i=2; i<=n; i++) {            if(max<dp[i]) {                max=dp[i];                end=i;            }        }        int sum=0;        for(i=end; i>=1; i--) {            sum+=a[i];            if(sum==max)  start=i; //此处不能break，因为题目要求是找第一个        }        cout<<"Case "<<count<<":"<<endl;        cout<<max<<" "<<start<<" "<<end<<endl;        if(count<t)            cout<<endl;    }    return 0;}

第二种代码更简洁

#include <iostream>using namespace std;int main() {    int t,count=0;    cin>>t;    for(count=1; count<=t; count++) {        int n,x,sum,max,start,end,nowstart;        cin>>n>>x;        sum=max=x;        start=end=nowstart=1;        for(int i=2; i<=n; i++) {            cin>>x;            if(sum+x<x) { //sum是指前i个数字的子串最大和  sum<0,更新                sum=x;                nowstart=i;            } else {                sum+=x;                //end=i; 此处不能记录end，因为// If there are more than one result, output the first one. Output a blank line between two cases.            }            if(sum>max) {                max=sum;                start=nowstart;                end=i;            }        }        cout<<"Case "<<count<<":"<<endl;        cout<<max<<" "<<start<<" "<<end<<endl;        if(count<t)            cout<<endl;    }    return 0;}