HDU2929 Bigger is Better[DP 打印方案 !]

Bigger is Better

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1106    Accepted Submission(s): 278

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

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题意：要用最多N根火柴摆出一个尽量大的正整数，而且这个数要能够被M整除

洛谷U5033

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一开始想d[i][j]表示i根火柴模m后为j的最大数字

然后想会溢出

然后想用高精怎么样

然后想高精慢，可以把d[i][j]的值分解成k*m+j，保存k和j就行了，结果并不好转移

然后搜了一下题解

 

用d[i][j]表示i根火柴拼成数字模m后为j有几位

初始化-1不可行，d[0][0]=0

用更新的写法，因为/10总感觉有点悬

d[i+ms[k]][(j*+k)%m]

这样dp完了之后再处理每一位是什么，用bit[i][j]表示，n倒着枚举
具体的思想就是，大到小枚举k，找[i][j]第一个能更新到的"合法的"[ti][tj]
对于d[i][j]==mxl，bit[i][j]=10代表这一位最高下面没有了

最后打印方案，从尾开始，因为每次加是加在最后

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=,M=,INF=1e9;
int n,m;
int ms[]={,,,,,,,,,},d[N][M],bit[N][M];
void dp(){
    int mxl=;
    memset(d,-,sizeof(d));
    d[][]=;
    for(int i=;i<=n;i++)
        for(int j=;j<m;j++) if(d[i][j]>=){
            for(int k=;k<=;k++) if(i+ms[k]<=n){
                if(d[i+ms[k]][(j*+k)%m]<d[i][j]+)
                    d[i+ms[k]][(j*+k)%m]=d[i][j]+;
                //if(i==0&&j==0) printf("o %d %d %d\n",i+ms[k],(j*10+k)%m,d[i+ms[k]][(j*10+k)%m]);
            }
            if(j==) mxl=max(mxl,d[i][j]);//printf("d %d %d %d\n",i,j,d[i][j]);
        }

    memset(bit,-,sizeof(bit));
    for(int i=n;i>=;i--)
        for(int j=;j<m;j++) if(d[i][j]>=){
            if(d[i][j]==mxl&&j==) {bit[i][j]=;continue;}
            for(int k=;k>=;k--) if(i+ms[k]<=n){
                int ti=i+ms[k],tj=(j*+k)%m;
                if(d[ti][tj]==d[i][j]+&&bit[ti][tj]>=) {bit[i][j]=k;break;}
            //    if(i==0&&j==0) printf("obit %d %d %d %d\n",ti,tj,d[ti][tj],bit[ti][tj]);
            }
        }
    //printf("hi %d %d %d\n",mxl,d[0][0],bit[0][0]);
    if(mxl>){
        int i=,j=,ti,tj;
        while(mxl-->){
            ti=i+ms[bit[i][j]];
            tj=(j*+bit[i][j])%m;
            printf("%d",bit[i][j]);
            i=ti;j=tj;
        }
    }else printf("-1");
    printf("\n");
}
int main(){int cas=;
    while(cin>>n){ if(n==) break;cin>>m;
        printf("Case %d: ",++cas);
        dp();
    }
}