CodeForces 215E Periodical Numbers 数位DP
2013-08-29 19:04
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题意:给你一个区间[l,r],求这个区间内满足条件的数,条件是:这个数的二进制表示时,dig[i] == dig[i+k],(0<k<len,且len%k==0,len为这个数的二进制代码长度)
思路:考虑[0,x]这个区间,若x的位数为len,当数的长度 i 为0~len-1时,则是无限制的,这时dp[i] = sum{2^(k-1)},k为满足条件的循环长度。而且还要去掉重复的,比如当长度为6时,循环长度为2,3的数均会重复计算,当数的长度为len时,则在限制下,计算满足条件的数,具体实现看代码注释:
思路:考虑[0,x]这个区间,若x的位数为len,当数的长度 i 为0~len-1时,则是无限制的,这时dp[i] = sum{2^(k-1)},k为满足条件的循环长度。而且还要去掉重复的,比如当长度为6时,循环长度为2,3的数均会重复计算,当数的长度为len时,则在限制下,计算满足条件的数,具体实现看代码注释:
#include <cstdio> #include <cstring> #include <string> #include <iostream> #include <map> #include <set> #include <vector> #include <cmath> #include <stack> #include <queue> #include <cstdlib> #include <algorithm> using namespace std; typedef __int64 int64; typedef long long ll; #define M 600005 #define N 1000005 #define max_inf 0x7f7f7f7f #define min_inf 0x80808080 #define mod 1000000007 #define lc rt<<1 #define rc rt<<1|1 ll dp[70] , r , l , table[70];//table[i] = 2^i; int dig[70]; ll Cal(int k)//计算长度为k的数,满足条件的个数 { int i , j; ll ret = 0; for (i = 1 ; i < k ; i++) { if (k%i)continue; dp[i] = table[i-1]; for (j = 1 ; j < i ; j++)//减掉重复计算的数 { if (i%j == 0) dp[i] -= dp[j]; } ret += dp[i]; } return ret; } ll Solve(ll k) { int i , j , len = 0; ll ret = 0 , temp = k , num; while (temp) { dig[++len] = temp&1; temp >>= 1; } //计算无限制时满足条件的数 for (i = 1 ; i < len ; i++)ret += Cal(i); for (i = 1 ; i < len ; i++)//长度为len时,枚举循环长度 { if (len%i)continue; num = 1; temp = 0; dp[i] = 0; for (j = len-1 ; j > len-i ; j--) { //若dig[j]==1则可以令dig[j]=0,转变成无限制的情况 if (dig[j])dp[i] += table[i-(len-j)-1]; num = num*2+dig[j]; } temp = num; int up = len/i; for (j = 1 ; j < up ; j++)num = (num<<i)+temp; //num保存的为循环长度为i,且循环内每一位都受限制的情况下的这个数 dp[i] += (num <= k);//若num比k小,则加入答案中 //去掉重复的计算的数 for (j = 1 ; j < i ; j++) { if (i%j == 0) dp[i] -= dp[j]; } ret += dp[i]; } return ret; } int main() { int i; for (table[0] = 1 , i = 1 ; i < 70 ; i++)table[i] = table[i-1]*2; while (~scanf("%I64d%I64d",&l,&r)) printf("%I64d\n",Solve(r)-Solve(l-1)); return 0; }
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