POJ3373：Changing Digits(记忆化)

Description

Given two positive integers n and k, you are asked to generate a new integer, say m, by changing some (maybe none) digits of n, such that the following properties holds:

1. m contains no leading zeros and has the same length as n (We consider zero itself a one-digit integer without leading zeros.)
2. m is divisible by k
3. among all numbers satisfying properties 1 and 2, m would be the one with least number of digits different from n
4. among all numbers satisfying properties 1, 2 and 3, m would be the smallest one

Input

There are multiple test cases for the input. Each test case consists of two lines, which contains n(1≤n≤10¹⁰⁰) and k(1≤k≤10⁴, k≤n) for each line. Both n and k will not contain leading zeros.

Output

Output one line for each test case containing the desired number m.

Sample Input

226191033219

Sample Output

2119103

看到一篇博客解释的非常详细，在此引用一下，传送门：http://blog.csdn.net/lyy289065406/article/details/6698787/

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char s[105];int dp[105][10],tem[105],a[105],mod,len,f[105][10005],k;void init(){    int i,j;    for(i = 0; i<=9; i++)        dp[1][i] = i%mod;    for(i = 2; i<=len; i++)        for(j = 0; j<=9; j++)            dp[i][j] = (dp[i-1][j]*10)%mod;    memset(f,0,sizeof(f));}bool dfs(int cnt,int l,int k)//从l开始，还要改变cnt个数字，使得余数k变为0{    int i,j;    if(!k)    {        for(i = 0; i<len; i++)            printf("%d",tem[i]);        printf("\n");        return 1;    }    if(!cnt) return 0;    if(l>len-1) return 0;    if(f[l][k]>=cnt) return 0;    for(i = l; i<len; i++)//高位开始，从小取    {        for(j = 0; j<a[i]; j++)        {            if(!i && !j)                continue;            tem[i] = j;            int temp = (k-dp[len-i][a[i]]+dp[len-i][j])%mod;            if(temp<0)                temp+=mod;            if(dfs(cnt-1,i+1,temp))                return 1;        }        tem[i] = a[i];    }    for(i = len-1; i>=l; i--)//低位开始，要变大    {        for(j = a[i]+1; j<=9; j++)        {            if(!i && !j)                continue;            tem[i] = j;            int temp = (k-dp[len-i][a[i]]+dp[len-i][j])%mod;            if(temp<0)                temp+=mod;            if(dfs(cnt-1,i+1,temp))                return 1;        }        tem[i] = a[i];    }    f[l][k] = cnt;//l开始取cnt个数不能使得k变为0    return 0;}int main(){    int i,j;    while(~scanf("%s",s))    {        scanf("%d",&mod);        len = strlen(s);        init();        k = 0;        for(i = 0; i<len; i++)            tem[i] = a[i] = s[i]-'0';       for(i = len-1; i>=0; i--)           k = (k+dp[len-i][a[i]])%mod;        for(i = 0; i<len; i++)            if(dfs(i,0,k))                break;    }    return 0;}