HDOJ 2929 Bigger is Better

DP。。。。好难的DP。。。

Bigger is Better

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 749    Accepted Submission(s): 190

Problem Description

Bob has n matches. He wants to compose numbers using the following scheme (that is, digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 needs 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 matches):

Write a program to make a non-negative integer which is a multiple of m. The integer should be as big as possible.

 

Input

The input consists of several test cases. Each case is described by two positive integers n (n ≤ 100) and m (m ≤ 3000), as described above. The last test case is followed by a single zero, which should not be processed.

 

Output

For each test case, print the case number and the biggest number that can be made. If there is no solution, output -1.Note that Bob don't have to use all his matches.

 

Sample Input

6 3
5 6
0

 

Sample Output

Case 1: 111
Case 2: -1

 

Source

2006 Asia Regional Xi-An

 

Recommend

lcy

 

 

#include <iostream> #include <cstdio> #include <cstring> using  namespace std; const  int MaxN= 120,MaxM= 3200,MOD= 100000000; const  int a[ 10]={ 6,  2,  5,  5,  4,  5,  6,  3,  7,  6}; typedef  int BIG[ 7]; BIG dp[MaxN][MaxM]; void BIG2B(BIG a,BIG b) {      for( int i= 0;i< 7;i++)         a =b; } bool BIGless(BIG a,BIG b) {     for(int i=6;i>=0;i--)     {         if(a<b) return true;         if(a>b) return false;     }     return false; } void BIGmultipe(BIG x,int k,BIG ret) {     for(int i=0;i<7;i++)         ret=x;     for(int i=0;i<7;i++)     {         ret=ret*10+k;         k=ret/MOD;         ret=ret%MOD;     } } void GET_DP(int n,int m) {     memset(dp,0,sizeof(dp));     for(int i=0;i<MaxN;i++) for(int j=0;j<MaxM;j++) dp[j][0]=-1;     dp[0][0][0]=0;     BIG ret,t;     memset(ret,0,sizeof(ret)); ret[0]=-1;     for(int i=0;i<n;i++)     {         for(int j=0;j<m;j++)         {             if(dp[j][0]==-1) continue;             for(int k=0;k<10;k++)             {                 if(a[k]+i>n) continue;                 BIGmultipe(dp[j],k,t);                 if(BIGless(dp[i+a[k]][(j*10+k)%m],t))                 {                     BIG2B(dp[i+a[k]][(j*10+k)%m],t);                     if((j*10+k)%m==0)                     {                         if(BIGless(ret,t))                             BIG2B(ret,t);                     }                 }             }         }     }     if(ret[0]==-1)     {         puts("-1");     }     else     {         int i;         for(i=6;i>=0;i--)             if(ret) break;         if(i==-1)         {             puts("0");         }         else         {             for(;i>=0;i--)             {                 printf("%d",ret);             }             putchar(10);         }     } } int main() {     int n,m,cas=1;     while(scanf("%d",&n)!=EOF&&n)     {         scanf("%d",&m);         printf("Case %d: ",cas++);         GET_DP(n,m);     }     return 0; }

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