• HDU


    B-number

    找出 ([1, X]) 中满足:

    1. 子串含 "13"
    2. 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;
    	}
    
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  • 原文地址:https://www.cnblogs.com/Tony-Double-Sky/p/9774709.html
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