编程艺术五 HASH表 0-1背包

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题目出处http://blog.csdn.net/v_JULY_v/article/details/6419466

O(N)时间复杂度 HASH表的用处得以体现

// FindSum.cpp : Defines the entry point for the console application.//#include "stdafx.h"#include "stdio.h"#define  N 6#define  M 1000int array[N]={1,2,4,7,11,15};int hash[M];int main(int argc, char* argv[]){int pos;int i;for(i = 0; i < N; i++)    //构建hash表{pos = array[i] % M;hash[pos]++;}int n;scanf("%d", &n);for(i = 0; i < N; i++){pos = (n - array[i]) % M;if (hash[pos]){if (pos == array[i] && hash[pos] > 1){printf("%d %d", array[i], pos);break;}else if (pos != array[i]){printf("%d %d", array[i], pos);break;}}}return 0;}

二.2010年中兴面试题
编程求解：
输入两个整数 n 和 m，从数列1，2，3.......n 中随意取几个数,

使其和等于 m ,要求将其中所有的可能组合列出来。

#include "stdio.h"int n;int m;int array[1000]={0};void Print(){int flag = 0;for(int i = 1; i < n; i++){if (!flag && array[i]){printf("%d",i);flag = 1;}else if (array[i])printf("+%d",i);}printf("\n",i);}void FindSum(int num, int sum)//m是当前的数，sum是和{if (num > n || sum > m)return;if (sum == m){Print();return;}array[num] = 1;FindSum(num+1, sum+num);  //取marray[num] = 0;FindSum(num+1, sum);    //不取m}void main(){printf("n:");scanf("%d", &n);printf("m:");scanf("%d", &m);FindSum(1, 0);}

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