标签:blog os io for div amp size log
由LIS的nlogn解法 可以得出最后统计数组中数的个数即为LIS的长度 这样就可以状压了
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
const int maxn = 20;
LL a,b,k;
int len = 0,lim[maxn];
LL f[maxn][1 << 10][11];
void getlim(LL n) {
len = 0;
memset(lim,0,sizeof(lim));
while(n) {
lim[len++] = n % 10; n /= 10;
}
}
int check(int state) {
int cnt = 0;
while(state) {
cnt += state & 1;
state >>= 1;
}
return cnt;
}
int gao(int state,int num) {
for(int i = num;i <= 9;i++) if(state & (1 << i)) {
state &= ~(1 << i);
break;
}
state |= (1 << num);
return state;
}
LL dfs(int now,int state,int first,int bound) {
if(now == 0) {
if(check(state) == k) {
return 1;
}
return 0;
}
LL ¬e = f[now][state][k];
int m = bound ? lim[now - 1] : 9;
if(!bound && first && note != -1) return note;
LL ret = 0;
for(int i = 0;i <= m;i++) {
int ns = gao(state,i);
if(i || first) ret += dfs(now - 1,ns,1,bound && i == m);
else ret += dfs(now - 1,state,0,bound && i == m);
}
if(!bound && first) note = ret;
return ret;
}
LL solve(LL num) {
getlim(num);
return dfs(len,0,0,1);
}
int main() {
memset(f,-1,sizeof(f));
int T,kase = 1;
cin >> T;
while(T--) {
cout << "Case #" << kase++ << ":" << " ";
cin >> a >> b >> k;
cout << solve(b) - solve(a - 1) << endl;
}
return 0;
}
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标签:blog os io for div amp size log
原文地址:http://www.cnblogs.com/rolight/p/3891132.html