uva12105 - Bigger is Better 大数

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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):

$\epsfbox{p3782.eps}$

Fig 1 Digits from 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 integersn (n $\le$ 100) andm (m $\le$ 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

N个火柴能组成被M整除的最大的数是多少。

dp[i][j]表示用i个火柴整除M余j的最大的数，v[k]表示组成数字k要几个火柴，dp[i+v[k]][(j*10+k)%M]=max(dp[i+v[k]][(j*10+k)%M],dp[i][j]*10+k)，我们一般循环算dp的时候都是从前面算出的状态来更新当前状态，而这个是拿当前状态去更新后面的状态，其实是一样的，到当前状态的时候它已经被前面所有能更新它的更新过。而且这个题这样做的好处在于i+v[k]个火柴的余数就是(j*10+k)%M，而如果i-v[k]的余数就不好算了。

这个题可能有50位数，要用大数。书上说可以不用大数，这个还真没想出来怎么弄。。

#include<cstring>#include<cstdio>#include<iostream>#include<climits>#include<cmath>#include<algorithm>#include<queue>#include<map>#define INF 0x3f3f3f3f#define MAXN 110#define MAXM 3010using namespace std;int N,M;int v[10]={6,2,5,5,4,5,6,3,7,6};char dp[MAXN][MAXM][60],ans[60],s[60];int compare(char *a,char *b){    int la=strlen(a),lb=strlen(b);    if(la>lb) return 1;    else if(la<lb) return -1;    return strcmp(a,b);}void DP(char *a,char *b,int k){    strcpy(s,b);    int l=strlen(s);    if(l==1&&s[0]=='0'){        s[l-1]='0'+k;        s[l]=0;    }    else{        s[l]='0'+k;        s[l+1]=0;    }    if(compare(s,a)>0) strcpy(a,s);}int main(){    //freopen("in.txt","r",stdin);    int cas=0;    while(scanf("%d",&N),N){        scanf("%d",&M);        memset(dp,0,sizeof(dp));        dp[0][0][0]='0';        for(int i=0;i<=N;i++)            for(int j=0;j<M;j++) if(strlen(dp[i][j])>0){                for(int k=0;k<10;k++) DP(dp[i+v[k]][(j*10+k)%M],dp[i][j],k);            }        ans[0]=0;        for(int i=N;i>0;i--) if(compare(ans,dp[i][0])<0) strcpy(ans,dp[i][0]);        printf("Case %d: ",++cas);        if(ans[0]==0) puts("-1");        else puts(ans);    }    return 0;}

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