2017-2018 ACM-ICPC Latin American Regional Programming Contest E - Enigma【DP】

题意：给一个由若干个‘?’和一些数字组成的数S，和一个数N，现在要求在？处填数，使其能够整除N并且值最小。

分析：用DP[i][j]表示后i位组成的数，能否构成模N为j的数。

然后枚举记录一个前缀的和对n取模，判断一下是否存在dp[i+1][n-mod]=1就可以了。

#include 
using namespace std;
char s[1004];
bool dp[1004][1004];
int base[1004];
int main(){
    int n;
    scanf("%s",s);
    int len = strlen(s);
    scanf("%d",&n);
    int sum = 1;
    dp[len][0]=1;
    base[0]=1;
    for (int i = 1; i <= len; ++i) {
        base[i] = base[i-1]*10%n;
    }
    for (int i = len - 1; i >= 0; --i) {
        if(s[i] == '?'){
            for (int j = 0; j <= 9; ++j) {
                if(i == 0 && j == 0)continue;
                int temp = sum * j % n;
                for (int k = 0; k < n; ++k) {
                    if(dp[i+1][k]){
                        dp[i][(k + temp)%n] = 1;
                    }
                }
            }
        }
        else {
            int temp = s[i] - '0';
            temp = temp * sum % n;
            for (int k = 0; k < n; ++k) {
                if(dp[i+1][k]){
                    dp[i][(k + temp)%n] = 1;
                }
            }
        }
        sum = sum * 10 % n;
    }
    sum = 0;
    bool yes = 1;
    for (int i = 0; i < len && yes; ++i) {
        if(s[i] == '?')
            for (int j = 0; j < 10; ++j) {
                if(i == 0 && j == 0)continue;
                int temp = sum + j*base[len - i - 1]%n;
                temp %= n;
                if(dp[i+1][(n - temp)%n]){
                    s[i] = j + '0';
                    break;
                }
            }
        if(s[i] == '?')yes = 0;
        else sum = (sum + (s[i]-'0')*base[len - i - 1]%n)%n;
    }
    if(yes)printf("%s",s);
    else printf("*\n");
}