Educational Codeforces Round 8 D. Magic Numbers

Posted 2020-06-14

Magic Numbers

题意:给定长度不超过2000的a,b;问有多少个x(a<=x<=b)使得x的偶数位为d,奇数位不为d;结果mod 1e9+7;

直接数位DP;注意在判断是否f[pos][mod] != -1之前，要判断是否为边界，否则会出现重复计算；

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define MS1(a) memset(a,-1,sizeof(a))
const int N = 2005,MOD = 1e9+7;
char a[N],b[N];
int f[N][N][2],n,m,d;
int dfs(char* s,int p,int mod,int on = 1)
{
    if(p == n) return mod == 0;
    int& ans = f[p][mod][on];
    if(ans != -1) return ans;
    ans = 0;
    for(int i = 0;i <= (on?s[p] - ‘0‘:9);i++){
        if(p%2 && i != d) continue;
        if(p%2 == 0 && i == d) continue;
        (ans += dfs(s,p+1,(mod*10+i)%m,on && i == s[p] - ‘0‘)) %= MOD;
    }
    return ans;
}
int calc(char* s)
{
    MS1(f);
    return dfs(s,0,0);
}
int main()
{
    scanf("%d%d",&m,&d);
    scanf("%s%s",a,b);
    n = strlen(a);
    for(int i = n-1;i >= 0;i--){
        if(a[i] == ‘0‘) a[i] = ‘9‘;
        else{a[i]--;break;}
    }
    printf("%d",(calc(b)-calc(a)+MOD)%MOD);
}

View Code

还有一种就是增加一维[2]，用来存储是否为边界；大同小异

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define MS1(a) memset(a,-1,sizeof(a))
const int N = 2005,MOD = 1e9+7;
char a[N],b[N];
int f[N][N][2],n,m,d;
int dfs(char* s,int p,int mod,int on = 1)
{
    if(p == n) return mod == 0;
    int& ans = f[p][mod][on];
    if(ans != -1) return ans;
    ans = 0;
    for(int i = 0;i <= (on?s[p] - ‘0‘:9);i++){
        if(p%2 && i != d) continue;
        if(p%2 == 0 && i == d) continue;
        (ans += dfs(s,p+1,(mod*10+i)%m,on && i == s[p] - ‘0‘)) %= MOD;
    }
    return ans;
}
int calc(char* s)
{
    MS1(f);
    return dfs(s,0,0);
}
int main()
{
    scanf("%d%d",&m,&d);
    scanf("%s%s",a,b);
    n = strlen(a);
    for(int i = n-1;i >= 0;i--){
        if(a[i] == ‘0‘) a[i] = ‘9‘;
        else{a[i]--;break;}
    }
    printf("%d",(calc(b)-calc(a)+MOD)%MOD);
}

View Code