joj 1197: Sum It Up
2010-03-09 08:41
302 查看
InputThe 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. 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 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. OutputFor 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 Input4 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 OutputSums 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 This problem is used for contest: 60 Submit / Problem List / Status / Discuss #include<stdio.h> int sum,n,total,res; int a[12]; int result[1000][12]; bool flag[12]; bool found; int count,count1; void sort(int n,int a[]) { int i,j,temp; for(i=0;i<n-1;i++) for(j=0;j<n-1-i;j++) if(a[j]<a[j+1]) { temp=a[j]; a[j]=a[j+1]; a[j+1]=temp; } } bool compare(int a[],int b[]) { int i; for(i=0;a[i]!=0&&b[i]!=0;i++) if(a[i]!=b[i]) return false; return true; } void Backtrack(int k) { int i; if(total==sum) { found=true; count1=0; for(i=0;i<k;i++) if(flag[i]) result[count][count1++]=a[i]; result[count][count1]=0; count++; return ; } if(k==n) { return ; } if((total+a[k])<=sum) { res-=a[k]; flag[k]=true; total+=a[k]; Backtrack(k+1); total-=a[k]; res+=a[k]; } if(total+res>=sum) { flag[k]=false; Backtrack(k+1); } } int main() { freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); while(scanf("%d%d",&sum,&n)!=EOF) { if(n==0) break; int i,j,k; total=0; res=0; count=0; found=false; for(i=0;i<n;i++) { flag[i]=false; scanf("%d",&a[i]); res+=a[i]; } sort(n,a); Backtrack(0); printf("Sums of %d:/n",sum); if(!found) printf("NONE/n"); else { for(i=0;i<count;i++) { for(j=0;j<=i;j++) { if(compare(result[j],result[i])) break; } if(j>=i) { printf("%d",result[i][0]); for(k=1;result[i][k]!=0;k++) printf("+%d",result[i][k]); printf("/n"); } } } } return 0; } |
相关文章推荐
- joj1197(深搜,Sum it up)
- 1197: Sum It Up
- 1711-Sum It Up 题解
- poj1564-Sum It Up(经典DFS)
- [置顶] hdu-1258 Sum It Up(dfs+去重)
- HDU 1258 Sum It Up
- HDU 1258 Sum It Up(DFS)
- zoj 1711 || poj 1564 Sum It Up(DFS~~~去重~)
- hdoj--1258--Sum It Up(dfs)
- HDU 1258 - Sum It Up
- 菜鸟学算法之POJ 1564 Sum It Up
- zoj 1711 Sum It Up
- HDU 1258 Sum It Up (dfs)
- 程序设计大赛—Sum It Up
- hdu 1258 Sum It Up 搜索
- HDU 1258 Sum it up 回溯法 暴力
- POJ 1564 Sum It Up (DFS+剪枝)
- HDU 1258 Sum It Up
- Sum It Up
- HDU 1268 Sum It Up【DFS】