Sum It Up Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4 Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → 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 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 Source 浙江工業大學第四屆大學生程序設計競賽 做練習的時候，糾結死啦！當時我一直在想怎樣判重，自己YY了一個hash函數 WA了。。。網上的一種解法不需要判重，剪枝就可以了。按照那個思路我重新敲了一下。code1: (dfs+剪枝)poj:Accepted 388K 0MS G++ 930B [cpp] #include <stdio.h> #include <string.h> int a[15]; int p[15]; int vis[15]; int t, n, flag; void dfs(int k, int sum) { int i; if(k>n || sum<0) return ; if(sum==0) { flag = 1; for(i=0; i<k-1; i++) printf("%d+",p[i]); printf("%d\n",p[i]); return ; } for(i=k; i<n; i++) if(!vis[i]) { if(sum-a[i]<0||(k>0&&a[i]>p[k-1])) continue; vis[i] = 1; p[k] = a[i]; dfs(k+1,sum-a[i]); vis[i] = 0; while(i+1<n&&a[i]==a[i+1]) i++; //搜索完畢後，若下一個搜索的數仍與當前相同，則尋找下一個不同的數進行搜索。｛去重｝ } } int main() { int i; while(scanf("%d%d",&t,&n),t+n) { for(i=0; i<n; i++) scanf("%d",&a[i]); i = 0; while(i<n&&a[i]>t) i++; printf("Sums of %d:\n",t); flag = 0; memset(vis,0,sizeof(vis)); dfs(i,t); if(!flag) printf("NONE\n"); } return 0; } #include <stdio.h> #include <string.h> int a[15]; int p[15]; int vis[15]; int t, n, flag; void dfs(int k, int sum) { int i; if(k>n || sum<0) return ; if(sum==0) { flag = 1; for(i=0; i<k-1; i++) printf("%d+",p[i]); printf("%d\n",p[i]); return ; } for(i=k; i<n; i++) if(!vis[i]) { if(sum-a[i]<0||(k>0&&a[i]>p[k-1])) continue; vis[i] = 1; p[k] = a[i]; dfs(k+1,sum-a[i]); vis[i] = 0; while(i+1<n&&a[i]==a[i+1]) i++; //搜索完畢後，若下一個搜索的數仍與當前相同，則尋找下一個不同的數進行搜索。｛去重｝ } } int main() { int i; while(scanf("%d%d",&t,&n),t+n) { for(i=0; i<n; i++) scanf("%d",&a[i]); i = 0; while(i<n&&a[i]>t) i++; printf("Sums of %d:\n",t); flag = 0; memset(vis,0,sizeof(vis)); dfs(i,t); if(!flag) printf("NONE\n"); } return 0; } code2:(用 set 去重：在POJ和ZOJ上提交全掛，不過hdu上能AC，呃呃呃~) HDU:Accepted 1258 0MS 340K 1286 B G++ [cpp] #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <functional> #include <vector> #include <string> #include <set> using namespace std; #define N 20 int t, n; int a[N]; int list[N]; bool vis[N]; set<string> s; bool flag; void dfs(int k, int sum) { int i; if(k>=n || sum<0) return; if(sum==0) { string str; for(i=0; i<k; i++) { str +=(list[i]/10) +'0'; str +=(list[i]%10) +'0'; } if(s.find(str)==s.end()) { s.insert(str); flag = 1; for(i=0; i<k-1; i++) printf("%d+",list[i]); printf("%d\n",list[i]); } return ; } for(i=k; i<n; i++) if(!vis[i]&&(k==0||a[i]<=list[k-1])) { vis[i] = 1; list[k] = a[i]; dfs(k+1,sum-a[i]); vis[i] = 0; } } int main() { int i; while(scanf("%d%d",&t,&n),t+n) { for(i=0; i<n; i++) { scanf("%d",&a[i]); } memset(vis,0,sizeof(vis)); printf("Sums of %d:\n",t); flag = false; s.clear(); dfs(0,t); if(!flag) printf("NONE\n"); } return 0; } #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <functional> #include <vector> #include <string> #include <set> using namespace std; #define N 20 int t, n; int a[N]; int list[N]; bool vis[N]; set<string> s; bool flag; void dfs(int k, int sum) { int i; if(k>=n || sum<0) return; if(sum==0) { string str; for(i=0; i<k; i++) { str +=(list[i]/10) +'0'; str +=(list[i]%10) +'0'; } if(s.find(str)==s.end()) { s.insert(str); flag = 1; for(i=0; i<k-1; i++) printf("%d+",list[i]); printf("%d\n",list[i]); } return ; } for(i=k; i<n; i++) if(!vis[i]&&(k==0||a[i]<=list[k-1])) { vis[i] = 1; list[k] = a[i]; dfs(k+1,sum-a[i]); vis[i] = 0; } } int main() { int i; while(scanf("%d%d",&t,&n),t+n) { for(i=0; i<n; i++) { scanf("%d",&a[i]); } memset(vis,0,sizeof(vis)); printf("Sums of %d:\n",t); flag = false; s.clear(); dfs(0,t); if(!flag) printf("NONE\n"); } return 0; }
