F(x)
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
题目大意：
给定多项式，对应的An表示对应的输入十进制数字的各个位数，
再给定数字A，B ，求在[0,B]区间内满足 f（i）<=f(A) 的i的个数。
解题思路：
参考：https://blog.csdn.net/wust_zzwh/article/details/52100392
代码：
#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
int a[12];
const int MAXN = 1e4+5;
int dp[12][MAXN];
int A,B;
int all;
int f(int x){
if(x==0) return 0;
int ans = f(x/10);
return ans*2+x%10;
}
int dfs(int pos,int sum,bool limit){
if(pos <= -1 && sum<=all) return 1;
if(sum>all) return 0;
if(!limit&&dp[pos][all - sum]!=-1) return dp[pos][all - sum];
int up = limit?a[pos]:9;
int res =0;
for(int i=0;i<=up;i++){
res += dfs(pos-1,sum+i*(1<<pos),limit && i==a[pos]);
}
if(!limit) dp[pos][all - sum] = res;
return res;
}
int slove(int input){
int pos = 0;
while(input>0){
a[pos++] = input%10;
input/=10;
}
int res = dfs(pos-1,0,true);
return res;
}
int main(){
int caseT;
scanf("%d",&caseT);
memset(dp,-1,sizeof(dp));
for(int i=1;i<=caseT;i++){
scanf("%d %d",&A,&B);
all = f(A);
int res = slove(B);
printf("Case #%d: %d\n",i,res);
}
return 0;
}