要理解这道题的代码得先知道LIS问题时间复杂度为O(nlogn)的算法思想,下面代码中get_news()函数有用到这种思想。
还不是太懂,先保存下,等明白了在详细解释一遍
#include <iostream>
#include <cstring>
using namespace std;
int t, k;
int bit[30];
long long dp[20][1 << 10][11];
int get_news(int x, int state)
{
for(int i = x; i < 10; ++ i)
{
if(state & (1 << i))
{
return (state ^ (1 << i)) | (1 << x);
}
}
return state | (1 << x);
}
int get_num(int state)
{
int cnt = 0;
while(state)
{
if(state & 1)
{
cnt++;
}
state >>= 1;
}
return cnt;
}
long long dfs(int pos, int state, bool limits, bool zeros)
{
if(pos == -1)
return get_num(state) == k;
if(!limits && dp[pos][state][k] != -1)
return dp[pos][state][k];
int en = limits ? bit[pos] : 9;
long long ans = 0;
for(int i = 0; i <= en; ++ i)
{
ans += dfs(pos - 1, (zeros && (i == 0)) ? 0 : get_news(i, state), (limits && (i == en)), zeros && (i == 0));
}
if(!limits)
{
dp[pos][state][k] = ans;
}
return ans;
}
long long solve(long long x)
{
int len = 0;
while(x)
{
bit[len++] = x % 10;
x /= 10;
}
return dfs(len - 1, 0, true, true);
}
int main()
{
cin >> t;
memset(dp, -1, sizeof(dp));
int ss = 1;
long long l, r;
while(t--)
{
cin >> l >> r >> k;
cout << "Case #" << ss++ << ": " << solve(r) - solve(l - 1) << endl;
}
return 0;
}