hdu 6205 card card card 最大子段和

#include<iostream>
#include<deque>
#include<memory.h>
#include<stdio.h>
#include<map>
#include<string>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#define INF 1<<29
using namespace std;
const int maxn = 1005000;

int A[maxn];
int B[maxn];
int C[maxn*2+1];//A,B数组的差，在尾部复制一遍，模拟循环
int D[maxn*2+1];//C数组的前缀和

int N;

inline void scan_d(int &ret){
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

int main (){

    while(~scanf("%d", &N)){

        memset(D,0,sizeof(D));

        for(int i=1;i<=N;i++)
            scan_d(A[i]);
        for(int i=1;i<=N;i++){
            scan_d(B[i]);
            
            C[i]=A[i]-B[i];
            C[N+i]=A[i]-B[i];
            D[i]=C[i]+D[i-1];
        }

        for(int i=N+1;i<=N+N;i++){
            D[i]=C[i-N]+D[i-1];
        }

        
        for(int i=1;i<=N;i++){
             for(int j=i;j<=i+N;j++){
                 if(D[j]-D[i-1]<0)//如果不能取了，即i~j的和变为负数了
                 {
                     i+=j-i;//把前面i堆放到后面去，然后继续判断
                     break;
                 }

                 //如果成功取完，输出i-1，即是答案
                 if(j==N){
                     printf("%d\n",i-1);
                     i=N;
                 }
             }
        }


    }
    return 0;
}

 

最大字段和 O(N)

//3d4-1 最大子段和问题的动态规划算法
#include "stdafx.h"
#include <iostream> 
using namespace std; 

int MaxSum(int n,int *a);

int main()
{
    int a[] = {-2,11,-4,13,-5,-2};

    for(int i=0; i<6; i++)
    {
        cout<<a[i]<<" ";
    }

    cout<<endl;
    cout<<"数组a的最大连续子段和为:"<<MaxSum(6,a)<<endl;

    return 0;
}

int MaxSum(int n,int *a)
{
    int sum=0,b=0;
    for(int i=1; i<=n; i++)
    {
        if(b>0)
        {
            b+=a[i];
        }
        else
        {
            b=a[i];
        }
        if(b>sum)
        {
            sum = b;
        }
    }
    return sum;
}