HOJ 1983 Beautiful numbers (数位dp)
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HOJ 1983 Beautiful numbers (数位dp)
链接:http://acm.hit.edu.cn/hoj/problem/view?id=1983
题意:求范围内的美丽数的个数,美丽数--------能被其每个数位上的数整除。
思路:和dp(一)有点类似,由于要是每位数的倍数,而各位数可能是0-9的任意一个,因此最坏情况是2-9所有数的公倍数(0,1不考虑),即2520.
每次对数位求取最小公倍数,看是否满足倍数情况,在数据上进行一定的hash处理,其他就都是在dp入门中讲过的模板了。
#include<stdio.h>#include<string.h>#include<iostream>using namespace std;long long hashh[2599]; // 转换数据类型的时候偷了下懒..直接用的ctrl+r 见谅..long long num[25];long long dp[25][2599][60];long long GCD(long long x,long long y) // 求最大公约数{ if(y == 0) return x; return GCD(y,x%y);}long long LCM(long long x,long long y) // 求最小公倍数{ return x/GCD(x,y)*y;}long long dfs(long long len,long long mod,long long lcm,bool limit)// len数值长度,mod对2520取模的值,lcm最小公倍数,limit前一位是否到达最大值{ long long sum = 0; if(len <= 0) return (mod%lcm)==0; if(!limit && dp[len][mod][hashh[lcm] ] != -1) return dp[len][mod][hashh[lcm] ]; long long over = limit?num[len]:9; for(int i = 0; i <= over; i++) { sum += dfs(len-1,(mod*10+i)%2520,i?LCM(lcm,i):lcm,(i==over)&&limit); // 每次取下一位都进行取模;算出最小公倍数 } if(!limit)dp[len][mod][hashh[lcm] ] = sum; return sum;}long long solve(long long number){ long long len = 0; while(number) { num[++len] = number%10; number /= 10; } return dfs(len,0,1,true);}void init(){ memset(dp,-1,sizeof(dp)); long long a = 0; for(int i = 1; i*i < 2520; i++) // 对2520的约数进行hash储存 { if(2520%i == 0) { hashh[i] = a++; hashh[2520/i] = a++; } }}int main(){ long long a,b; init(); while(~scanf("%lld%lld",&a,&b)) { printf("%lld\n",solve(b)-solve(a-1)); } return 0;}
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