F(x)
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8478 Accepted Submission(s): 3339
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
2013 ACM/ICPC Asia Regional Chengdu Online
#include <bits/stdc++.h>
using namespace std;
int a[15], b[15];
int dp[15][25000];
int dfs(int pos, int sum, bool lim)
{
if (pos == 0)
return 1;
if (!lim && dp[pos][sum])
return dp[pos][sum];
int res = 0;
int up = lim ? b[pos] : 9;
for (int i = 0; i <= up; i++)
{
int t = sum - i * (1 << (pos - 1));
if (t < 0)
break;
res += dfs(pos - 1, t, lim && (i == up));
}
if (!lim)
dp[pos][sum] = res;
return res;
}
int solve(int x, int y)
{
// memset(dp, 0, sizeof dp);
/// 500ms T 1w 不能每次清空
int LIMIT = 0;
int w1 = 0, w2 = 0;
while (x > 0)
{
a[++w1] = x % 10;
x /= 10;
}
for (int i = 1; i <= w1; i++)
LIMIT += a[i] * (1 << (i - 1));
while (y > 0)
{
b[++w2] = y % 10;
y /= 10;
}
return dfs(w2, LIMIT, 1);
/// 低位允许的总和
}
int main()
{
int t;
scanf("%d", &t);
for (int i = 1; i <= t; i++)
{
int a, b;
scanf("%d %d", &a, &b);
printf("Case #%d: %d\n", i, solve(a, b));
}
return 0;
}