题意
T测试数据,每组求[0,B]区间内有多少个数x的F(x)小于F(A),
假设A为一个n位数(),F(A)定义为:
思路
依旧记忆化搜索的思路,定义dp[pos][sum]表示选到pos位,后面的F值加和不超过sum的情况。
#include<bits/stdc++.h>
using namespace std;
int dp[10][5000], FA;
string num;
int dfs(int pos, int sum, bool limit)
{
if (pos == -1) return sum >= 0;
if (sum < 0) return 0;
if (!limit && dp[pos][sum] != -1) return dp[pos][sum];
int E = limit?num[pos]-'0':9, ans = 0;
for (int i = 0; i <= E; i++)
ans += dfs(pos-1, sum-i*(1<<pos), limit&&(i==E));
if (!limit) dp[pos][sum] = ans;
return ans;
}
int solve(int x)
{
num = to_string(x); reverse(num.begin(), num.end());
return dfs(num.size()-1, FA, 1);
}
int main()
{
memset(dp, -1, sizeof(dp));
int T, CASE = 1; scanf("%d", &T);
while (T--)
{
int A, B; scanf("%d%d", &A, &B);
FA = 0;
num = to_string(A); reverse(num.begin(), num.end());
for (int i = num.size()-1; i >= 0; i--)
FA += (num[i]-'0')*(1<<i);
printf("Case #%d: %d\n", CASE++, solve(B));
}
return 0;
}
/*
3
0 100
1 10
5 100
*/