You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
|
|
|
|
#include <cstdio>
|
|
|
|
|
#include <cstring>
|
|
|
|
|
#include <iostream>
|
|
|
|
|
#include <cmath>
|
|
|
|
|
|
|
|
|
|
using namespace std;
|
|
|
|
|
const int N = 32; //数位的长度上限
|
|
|
|
|
const int M = 1024 * 10 + 10; //剩余的数大小(状态集合按这个划分)
|
|
|
|
|
int f[N][M]; //结果DP数组
|
|
|
|
|
int a[N]; //数位拆分出来的数组
|
|
|
|
|
int fa;
|
|
|
|
|
//利用递归计算出
|
|
|
|
|
int F(int x) {
|
|
|
|
|
if (x == 0) return 0;
|
|
|
|
|
return F(x / 10) * 2 + x % 10;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/**
|
|
|
|
|
* @param pos 数位位置
|
|
|
|
|
* @param st 剩余对比值,初始值是F(x)
|
|
|
|
|
* @param limit 是不是贴上界
|
|
|
|
|
* @return
|
|
|
|
|
*/
|
|
|
|
|
int dfs(int pos, int st, bool limit) {
|
|
|
|
|
if (pos == 0) return st >= 0; //如果到达最后,并且有剩余,表示f(i)<= f(x),计数++
|
|
|
|
|
if (st < 0) return 0; //中途或最后一旦发生小于0情况,剪枝
|
|
|
|
|
if (!limit && ~f[pos][st]) return f[pos][st];
|
|
|
|
|
int ans = 0;
|
|
|
|
|
int up = limit ? a[pos] : 9;
|
|
|
|
|
for (int i = 0; i <= up; i++)
|
|
|
|
|
//st = st - i * 2^ ( p-1 )
|
|
|
|
|
ans += dfs(pos - 1, st - i * (1 << (pos - 1)), limit && i == a[pos]);
|
|
|
|
|
|
|
|
|
|
if (!limit) f[pos][st] = ans;
|
|
|
|
|
return ans;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
inline int calc(int x) {
|
|
|
|
|
int al = 0;
|
|
|
|
|
while (x) a[++al] = x % 10, x /= 10;
|
|
|
|
|
return dfs(al, fa, true);
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
int main() {
|
|
|
|
|
//不加78MS,加了31MS,效果还是很明显的
|
|
|
|
|
ios::sync_with_stdio(false);
|
|
|
|
|
cin.tie(0);
|
|
|
|
|
|
|
|
|
|
int T;
|
|
|
|
|
int cnt = 1;
|
|
|
|
|
cin >> T; //这个数值太BT了,最大10000次!
|
|
|
|
|
|
|
|
|
|
//优化的写法
|
|
|
|
|
memset(f, -1, sizeof f);
|
|
|
|
|
while (T--) {
|
|
|
|
|
int x, y;
|
|
|
|
|
cin >> x >> y;
|
|
|
|
|
fa = F(x); //计算出F(x)值,这是一个固定值
|
|
|
|
|
printf("Case #%d: %d\n", cnt++, calc(y));
|
|
|
|
|
}
|
|
|
|
|
return 0;
|
|
|
|
|
}
|