Codeforces 628D Magic Numbers

题意:

求在 [a,b] a,b 不含前导0)中的 dmagic 数中有多少个是 m 的倍数。

分析:

计数dp

Let’s call a number d-magic if digit d appears in decimal presentation of the number on even positions and nowhere else.

仔细读题并观察例子就可以明确 dmagic 从左到右所有偶数位置上都是d,奇数位上不能是d
[a,b] ,即可转化为求 [1,a],[1,b] 中的满足条件的数,最后相减,注意判断 a 是否满足条件。
dp[i][j][k] 表示前缀为 i 位,余数为 j ,等于(1)或者小于(0)上限的方案数。容易想到状态转移的过程,注意分填入的数字小于和等于上限对应元素两种情况处理。

代码:

#include<cstdio>
#include<cstring>
const int mod = 1e9+7, maxn = 2005;
int m, d;
int dp[maxn][maxn][2];
char a[maxn], b[maxn];
int num[maxn];
int solve(char* A)
{
    memset(dp, 0, sizeof(dp));
    memset(num, 0, sizeof(num));
    int cnt = strlen(A);
    for(int i = 0; i < cnt; i++){
        num[i+1] = A[i] - '0';
    }
    for(int i = 1; i<=num[1];i++){
        if(i == d)continue;
        if(i < num[1]){
            dp[1][i%m][0]++;
        }else{
            dp[1][i%m][1]++;
        }
    }

    for(int i = 2; i <= cnt; i++){
        for(int j = 0; j < m; j++){
            if(i%2==0){
                dp[i][(j * 10 + d)%m][0] = (dp[i][(j * 10 + d)%m][0] + dp[i - 1][j][0])%mod;
                if(d < num[i])
                    dp[i][(j * 10 + d)%m][0] = ( dp[i][(j * 10 + d)%m][0] + dp[i - 1][j][1])%mod;
                else if(d == num[i])
                    dp[i][(j * 10 + d)%m][1] = ( dp[i][(j * 10 + d)%m][1] + dp[i-1][j][1])%mod;
            }else{
                for(int k = 0; k < 10; k++){
                    if(k == d) continue;
                        dp[i][(j * 10 + k)%m][0] = (dp[i][(j * 10 + k)%m][0] + dp[i - 1][j][0])%mod;
                    if(k < num[i])
                        dp[i][(j * 10 + k)%m][0] = ( dp[i][(j * 10 + k)%m][0] + dp[i - 1][j][1])%mod;
                    else if(k == num[i])
                        dp[i][(j * 10 + k)%m][1] = ( dp[i][(j * 10 + k)%m][1] + dp[i-1][j][1])%mod;
                    }
                }
            }
        }

    return (dp[cnt][0][0] + dp[cnt][0][1])%mod;

}
int is(char* a)
{
    int res = 0;
    for(int i = 0; i < strlen(a); i++){
        int a1 = a[i] - '0';
        if(i%2 == 0){
            if(a1 == d) return 0;
        }
        if(i%2==1){
            if(a1 != d)  return 0;
        }
        res = (res*10 +a1)%m;
    }
    return res == 0;
}
int main (void)
{
    scanf("%d%d",&m,&d);
    scanf("%s%s", a, b);
    printf("%d\n", (solve(b) - solve(a) +is(a)+mod)%mod) ;
    return 0;
}

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