HDOJ 1258 Sum It Up(DFS)

Sum It Up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 218 Accepted Submission(s): 110

Problem Description

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 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 x1,...,xn. 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 x1,...,xn 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 distince; the same sum connot appear twice.

 

Sample Input

4 6 4 3 2 2 1 15 3 2 1 1400 12 50 50 50 50 50 50 25 25 25 25 25 250 0

 

Sample Output

Sums of 4:43+12+22+1+1Sums of 5:NONESums of 400:50+50+50+50+50+50+25+25+25+2550+50+50+50+50+25+25+25+25+25+25

 

Source

浙江工业大学第四届大学生程序设计竞赛

 

Recommend

JGShining

 

这个去重我下了很久

#include<iostream>  #include<cstdio>  #include<cstring>  #include<algorithm>  #include<cmath>  using namespace std;int t;int n;int a[15];int ans[15];//答案int ans2[15];//存前一次的结果bool flag;//判断是否找到bool flag2;//判断是否重复bool cmp(int a, int b){return a > b;}void dfs(int index,int sum,int deep)//下标  总和  深度{if (sum == t)//找到{if (!flag) //第一次找到{for (int i = 0; i < deep - 1; i++)//输出结果{printf("%d+", ans[i]);ans2[i] = ans[i];//更新数组}ans2[deep - 1] = ans[deep - 1];printf("%d\n", ans[deep - 1]);flag = true;return;}else {for (int i = 0; i < deep; i++)//结果是否重复{if (ans2[i] != ans[i]) { flag2 = true; break; }}if (flag2)//没有重复{for (int i = 0; i < deep - 1; i++){printf("%d+", ans[i]);ans2[i] = ans[i];}ans2[deep - 1] = ans[deep - 1];//更新printf("%d\n", ans[deep - 1]);flag2 = false;//还原return;}}}for (int i = index+1; i < n; i++)//从前往后按顺序找所以i是这个值{if (a[i] + sum > t){ while (a[i+1]==a[i])//重复的不用走了 剪枝{i++;}continue; }ans[deep] = a[i];dfs(i, sum + a[i],deep+1);//小于等于t}}int main(){while (scanf("%d%d", &t, &n) != EOF){if (t == 0 && n == 0)break;for (int i = 0; i < n; i++){scanf("%d", &a[i]);}sort(a, a + n, cmp);//从大到小排序 才能得到想要的结果flag = false;printf("Sums of %d:\n", t);memset(ans, 0, sizeof(ans));for (int i = 0; i < n; i++){flag2 = false;if (i > 0 && a[i] == a[i - 1])continue;//开头的数字不能重复ans[0] = a[i];dfs(i, a[i], 1);}if (!flag)printf("NONE\n");}return 0;}

结果看了别人的题解。

有个大佬的做法很简洁，代码：

#include <stdio.h>  #include <string.h>  #include <algorithm>  using namespace std;    int n,len,a[20],b[20],cnt;    int cmp(int a,int b)  {      return a>b;  }    void dfs(int x,int posa,int sum,int posb)  {      int i;      if(sum>n)          return;      if(sum == n)      {          cnt++;          for(i = 0; i<posb; i++)          {              if(i)                  printf("+%d",b[i]);              else                  printf("%d",b[i]);          }          printf("\n");      }      for(i = posa; i<len; i++)      {          b[posb] = a[i];          dfs(a[i],i+1,sum+a[i],posb+1);          while(i+1<len && a[i] == a[i+1])//搜索完毕后，若下一个搜索的数仍与当前相同，则跳过直至不相同                i++;      }  }    int main()  {      int i;      while(~scanf("%d%d",&n,&len),n+len!=0)      {          for(i = 0; i<len; i++)              scanf("%d",&a[i]);          sort(a,a+len,cmp);          printf("Sums of %d:\n",n);          cnt = 0;          dfs(0,0,0,0);          if(!cnt)              printf("NONE\n");      }        return 0;  }