F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5055    Accepted Submission(s): 1880 Problem Description For a decimal number x with n digits (A _nA _n-1A _n-2 ... A ₂A ₁), we define its weight as F(x) = A _n * 2 ^n-1 + A _n-1 * 2 ^n-2 + ... + A ₂ * 2 + A ₁ * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).   Input The first line has a number T (T <= 10000) , indicating the number of test cases. For each test case, there are two numbers A and B (0 <= A,B < 10 ⁹)   Output For every case,you should output "Case #t: " at first, without quotes. The t is the case number starting from 1. Then output the answer.   Sample Input 3 0 100 1 10 5 100   Sample Output Case #1: 1 Case #2: 2 Case #3: 13

题意：找出i在0到b之间的f(i)小于等于f(a)的数的个数。

思路：数位dp。主要在于状态转移不好想。dp[i][j]表示i位数比j小的数的个数。用递归完成的话就只需要思考边界和状态转移。

边界：

dp[i][j]如果j小于0，显然是dp[i][j]=0的，如果i==0，说明就是0，显然任何数都比0大，所以dp[i][j]对于j>=0的时候dp[i][j]=1,否则dp[i][j]=0。

状态转移：

dp[i][j]+=dp[i-1][j-k*(1<<(i))]；j的初始值是sum

完成上述两步推导就能开始写这题了。

#include <bits/stdc++.h> using namespace std; typedef long long LL; const int MOD = 1000000007; const int maxn = 200010; int num[30],len; LL dp[30][10000]; int a[50]; int sum; LL dfs(int pos,LL zhi,int last)//zhi是差值 { if(pos<0) return zhi>=0;//当差值大于等于0时，说明这个数符合要求，+1，否则+0 if(zhi<0) return 0;//当差值小于0时就+0， if(!last&&dp[pos][zhi]!=-1) { return dp[pos][zhi]; } int len=last?num[pos]:9; LL res = 0; for(int i=0; i<=len; i++) { res += dfs(pos-1,(zhi-i*(1<<pos)),last&&(i==len)); } if(!last)dp[pos][zhi]= res; return res; } LL solve(LL n) { len = 0; while(n) { num[len++] = n; n /= 10; } return dfs(len-1,sum,1);//用sum一直去减 } int main() { memset(dp,-1,sizeof(dp)); int _; scanf("%d",&_); int g=0; while(_--) { LL n,m; ++g; scanf("%I64d%I64d",&m,&n); sum=0; int l=0; while(m)//算sum { sum+=(m)*(1<<l); l++; m=m/10; } printf("Case #%d: ",g); if(sum==0) printf("%d\n",1); else printf("%I64d\n",solve(n)); } return 0; }