深搜与不重复组合

zoj  1711

Sum It Up

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t = 4, n = 6, and the list is [4, 3, 2, 2, 1, 1], then there are four different sums that equal 4: 4, 3+1, 2+2, and 2+1+1. (A number can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.

Input

The input file will contain one or more test cases, one per line. Each test case contains t, the total, followed by n, the number of integers in the list, followed by n integers x 1 , . . . , x n . If n = 0 it signals the end of the input; otherwise, t will be a positive integer less than 1000, n will be an integer between 1 and 12 (inclusive), and x 1 , . . . , x n will be positive integers less than 100. All numbers will be separated by exactly one space. The numbers in each list appear in nonincreasing order, and there may be repetitions.

Output

For each test case, first output a line containing `Sums of', the total, and a colon. Then output each sum, one per line; if there are no sums, output the line `NONE'. The numbers within each sum must appear in nonincreasing order. A number may be repeated in the sum as many times as it was repeated in the original list. The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum. In other words, the sums must be sorted by their first number; sums with the same first number must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on. Within each test case, all sums must be distinct; the same sum cannot appear twice.

Sample Input

4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25
0 0

Sample Output

Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25

题意：给出最多 12 个数，让你任意选择其中若干数，要求其和为 t 的序列，

这题就是组合数学里的不重复组合，

不重复组合 

 就是这种组合的模型。。。。。。
输入 n 个数，求这 n 个数构成的集合的所有子集，不允许输出重复的项。 
输入样例 
3 
1 1 3 
输出样例 
1 
1 1 
1 1 3 
1 3 
3 

Accepted1711C++0188我的代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <stdio.h>#define MAX_N 10int n, m,t;       //输入 n 个数，其中本质不同的有 m 个int rcd[MAX_N]; //记录每个位置填的数int used[MAX_N];//标记 m 个数可以使用的次数int num[MAX_N]; //存放输入中本质不同的 m 个数bool flag;void unrepeat_combination(int l, int p,int ans){ int i; if(ans==t) {     flag=1;     printf("%d", rcd[0]);     for (i=1; i<l; i++)     {    printf("+%d", rcd[i]);     }    printf("\n");    return; } else if(ans>t) return; for (i=p; i<m; i++)  //循环依旧从 p 开始，枚举剩下的本质不同if(used[i] > 0){//若还可以用, 则可用次数减 used[i]--; rcd[l] = num[i];   //在 l 位置放上该unrepeat_combination(l+1, i,ans+num[i]); //填下一个位置 used[i]++; //可用次数恢复 }}int main(){ int i, j, val; while(scanf("%d%d",&t,&n),t&&n) {     flag=0;     m = 0; for (i=0; i<n; i++){ scanf("%d", &val); for (j=0; j<m; j++) if (num[j] == val){ used[j]++; break; } if (j == m){ num[m] = val; used[m++] = 1; } } printf("Sums of %d:\n",t); unrepeat_combination(0,0,0); if(flag==0) printf("NONE\n"); } return 0;}