Poj1416-shredding company

 

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Question:

The company wants to invent a new shredder that requires the new shredder to cut the number on the paper into the nearest value without exceeding the target value. For example, if the value of target is 50, and the number on the paper is 12346, you should cut the number into four parts: 1, 2, 34, and 6. Because the sum of 43 (= 1 + 2 + 34 + 6) obtained in this way is the closest possible and not more than 50. (For example, 1, 23, 4, and 6 are not allowed, because their sum is not as close as 43 to 50, and 12, 34, and 6 are not allowed, because their sum exceeds 50. There are three requirements for broken paper:

1. If the target value is equal to the value on the paper, it cannot be switched.
2. If there is no way to cut the number on the paper into smaller than target, an error is output. For example, if the value of target is 1 and the number on the paper is 123, the sum you get is greater than 1.
3. If more than one cut method is used to obtain the optimal value, the rejected is output. For example, if the value of target is 15 and the number on the paper is 111, the following two cut methods are available: 11, 1, or 11.
Your task is to write a program to divide the numbers for the best value.

Solution:

Use the brute force DFS search to list all the situations. For example, 111 is used to first split 111 into, 1, and the second step into, and the third step, in this way, all the search results will be completed, and only the output can be judged during statistics.

Code:

#include <iostream>using namespace std;int dis[1000000];int maxn,sum,v[10],vt[10],len,t;char num[10];void dfs(int x,int m,int cap){int i,y,flag;if(x>len-1){flag=1;for(i=0;i<cap;i++)if(vt[i]!=v[i]) flag=0;if(maxn<m){maxn=m;sum=cap;for(i=0;i<cap;i++) vt[i]=v[i];}if(flag==0) dis[m]++;return;}y=0;for(i=x;i<len;i++){y=y*10+(num[i]-'0');if(y+m<=t){v[cap]=y;dfs(i+1,y+m,cap+1);}else break;}}int main(){int s,i;while(scanf("%d%s",&t,num)){memset(v,0,sizeof(v));memset(vt,0,sizeof(vt));memset(dis,0,sizeof(dis));len = strlen(num);if(t==0 && len==1 && num[0]=='0') break;s=0;for(i=0;i<len;i++)s+=num[i]-'0';if(s>t) {printf("error\n");continue;}maxn=0,sum=0;dfs(0,0,0);if(dis[maxn]!=1)printf("rejected");else{printf("%d",maxn);for(i=0;i<sum;i++) printf(" %d", vt[i]);}printf("\n");}return 0;}