# F(x)

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 382    Accepted Submission(s): 137

Problem Description
For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2 * 2 + A1 * 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 < 109)

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

Source

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dp[i][j]表示i位值<=j 的总数

/* ***********************************************
Author        :kuangbin
Created Time  :2013/9/14 星期六 12:45:42
File Name     :2013成都网络赛\1007.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

int dp[20][200000];

int bit[20];

int dfs(int pos,int num,bool flag)
{
if(pos == -1)return num >= 0;
if(num < 0)return 0;
if(!flag && dp[pos][num] != -1)
return dp[pos][num];
int ans = 0;
int end = flag?bit[pos]:9;
for(int i = 0;i <= end;i++)
{

ans += dfs(pos-1,num - i*(1<<pos),flag && i==end);
}
if(!flag)dp[pos][num] = ans;
return ans;
}

int F(int x)
{
int ret = 0;
int len = 0;
while(x)
{
ret += (x%10)*(1<<len);
len++;
x /= 10;
}
return ret;
}
int A,B;
int calc()
{
int len = 0;
while(B)
{
bit[len++] = B%10;
B/=10;
//cout<<bit[len-1]<<endl;
}
//cout<<F(A)<<endl;
return dfs(len-1,F(A),1);
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int iCase = 0;
scanf("%d",&T);
memset(dp,-1,sizeof(dp));
while(T--)
{
iCase++;
scanf("%d%d",&A,&B);
printf("Case #%d: %d\n",iCase,calc());
}
return 0;
}

posted on 2013-09-15 00:25 kuangbin 阅读(...) 评论(...) 编辑 收藏

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