记录编号 |
584694 |
评测结果 |
AAAAAAAAAAAAAAAA |
题目名称 |
魔药 |
最终得分 |
100 |
用户昵称 |
┭┮﹏┭┮ |
是否通过 |
通过 |
代码语言 |
C++ |
运行时间 |
0.111 s |
提交时间 |
2023-11-14 17:11:14 |
内存使用 |
7.16 MiB |
显示代码纯文本
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const long long N = 1e6+10,mod = 1e9+7;
int n,m,k,p = 1e9+7;
ll s[N] = {1,1},ans;
char c[N];
ll pow(ll x,int y){
if(y == 0)return 1;
ll ans = pow(x,y/2);
if(y % 2)return (ans * ans) % mod * x % mod;
else return (ans * ans) % mod;
}
ll C(int x,int y){
return (s[x] * pow(s[x-y],p-2)) % mod * pow(s[y],p-2) % mod;
}
int main(){
freopen("sleeping.in","r",stdin);
freopen("sleeping.out","w",stdout);
scanf("%lld%lld%s",&n,&m,c+1);
for(int i = 1;i <= n;i++)if(c[i] == '1')k++;
for(int i = 2;i <= n;i++)s[i] = s[i-1] * i % mod;
if(m > n){
printf("0 %lld\n",C(n,k));
return 0;
}
if(m % 2 == 0)ans = 2 * C(k-1,k-m/2) * C(n-k-1,k-m/2) % mod;
else ans = (C(k-1,k-m/2) * C(n-k-1,n-k-m/2-1) % mod + (C(k-1,k-m/2-1) * C(n-k-1,n-k-m/2))) % mod;
// cout<<ans<<endl;
long long s = C(n,k) - ans;
if(s < 0)s += mod;
printf("%lld %lld\n",ans,s);
return 0;
}