B-number
找出 ([1, X]) 中满足:
- 子串含 "13"
- 能被 (13) 整除的数
的个数
错误日志: (B == 1) 时也可以转移到 (1) (没有分析清楚转移)
Solution
数位(dp)
满足两个性质: 能被 (13) 整除 , 子串含 (13)
状态设计: (dp[maxn][0/1/2][0 - 12]) 两个状态,
状态一
(0 -->) 无特殊性质
(1 -->) 数以 (1) 结尾
(2 -->) 数中含 ("13") 子串
状态二
次数对 (13) 取模的结果
状态二比较好转移, 状态一转移如下:
LL cmd = 0;
if((B == 0 || B == 1) && i == 1)cmd = 1;
else if(B == 1 && i == 3)cmd = 2;
else if(B == 2)cmd = 2;
Code
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
#define REP(i, x, y) for(LL i = (x);i <= (y);i++)
using namespace std;
LL RD(){
LL out = 0,flag = 1;char c = getchar();
while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
return flag * out;
}
const LL maxn = 19;
LL num[maxn];
LL dp[maxn][3][maxn];//数位,是否含13,除13余数多少
LL DP(LL Index, LL B, LL mod, bool limit){
if(Index == 0)return ((B == 2) && (mod == 0));
if(!limit && dp[Index][B][mod] != -1)return dp[Index][B][mod];
LL ans = 0, up = limit ? num[Index] : 9;
REP(i, 0, up){
LL cmd = 0;
if((B == 0 || B == 1) && i == 1)cmd = 1;
else if(B == 1 && i == 3)cmd = 2;
else if(B == 2)cmd = 2;
ans += DP(Index - 1, cmd, ((mod * 10 + i) % 13 + 13) % 13, limit && i == num[Index]);
}
if(!limit)dp[Index][B][mod] = ans;
return ans;
}
LL solve(LL x){
LL len = 0;
while(x){
num[++len] = x % 10;
x /= 10;
}
return DP(len, 0, 0, 1);
}
int main(){
memset(dp, -1, sizeof(dp));
LL x;
while(scanf("%lld", &x) != EOF){
printf("%lld
", solve(x));
}
return 0;
}