2017 Multi-University Training Contest - Team 8 1011& HDU6143 Killer Names(容斥|| 第二类斯特林数)
2017-08-17 20:53
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题意:给你m种字符,让你填充两个长度为n的字符串,要求两个字符串不能有相同的字符,问最多有多少种填法。
n,m<=2000
思路:解这题的关键是解决用恰好k个字符去填充长度为n的字符有多少种方案。比赛时知道容斥可以做,但没想出
来。。所以用了第二类斯特林去做,第二类斯特林数(点击打开链接):把p元素集合划分到k个不可区分的盒子里且
没有空盒子的划分个数。在这就是看作n个位置填充k个颜色,因为颜色可区分,所以要成k的阶乘。
容斥做法:
第二类斯特林代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const int maxn = 2e3+5;
ll s[maxn][maxn], c[maxn][maxn], fac[maxn];
void init()
{
memset(s, 0, sizeof(s));
s[1][1] = 1;
for(int i = 2; i < maxn; i++)
for(int j = 1; j <= i; j++)
s[i][j] = (s[i-1][j-1]+(ll)j*s[i-1][j])%mod;
fac[0] = fac[1] = 1;
for(int i = 2; i < maxn; i++)
fac[i] = fac[i-1]*i%mod;
for(int i = 1; i < maxn; i++)
for(int j = 0; j <= i; j++)
{
if(i == j || !j) c[i][j] = 1;
else c[i][j] = (c[i-1][j]+c[i-1][j-1])%mod;
}
}
ll qpow(ll p, int q)
{
ll ans = 1;
while(q)
{
if(q%2) ans = ans*p%mod;
p = p*p%mod;
q /= 2;
}
return ans;
}
int main(void)
{
init();
int n, m, t;
cin >> t;
while(t--)
{
scanf("%d%d", &n, &m);
ll ans = 0;
int e = min(n, m-1);
for(int i = 1; i <= e; i++)
{
ans = (ans + s
[i]*fac[i]%mod*qpow(m-i, n)%mod*c[m][i]%mod)%mod;
}
printf("%lld\n", ans%mod);
}
return 0;
}
容斥代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e3+5;
const int mod = 1e9+7;
typedef long long ll;
ll c[maxn][maxn], rec[maxn], fac[maxn];
void init()
{
fac[0] = fac[1] = 1;
for(int i = 2; i < maxn; i++)
fac[i] = fac[i-1]*i%mod;
for(int i = 1; i < maxn; i++)
for(int j = 0; j <= i; j++)
if(!j || i==j) c[i][j] = 1;
else c[i][j] = (c[i-1][j]+c[i-1][j-1])%mod;
}
ll qmod(ll x, int p)
{
ll ans = 1;
while(p)
{
if(p%2) ans = ans*x%mod;
x = x*x%mod;
p /= 2;
}
return ans;
}
int main(void)
{
init();
int t, n, m;
cin >> t;
while(t--)
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++)
{
rec[i] = qmod(i, n);
for(int j = 1; j < i; j++)
rec[i] = (rec[i]-c[i][j]*rec[j]%mod+mod)%mod;
}
ll ans = 0;
for(int i = 1; i <= min(n, m-1); i++)
ans = (ans+c[m][i]*rec[i]%mod*qmod(m-i, n)%mod)%mod;
printf("%lld\n", ans);
}
return 0;
}
n,m<=2000
思路:解这题的关键是解决用恰好k个字符去填充长度为n的字符有多少种方案。比赛时知道容斥可以做,但没想出
来。。所以用了第二类斯特林去做,第二类斯特林数(点击打开链接):把p元素集合划分到k个不可区分的盒子里且
没有空盒子的划分个数。在这就是看作n个位置填充k个颜色,因为颜色可区分,所以要成k的阶乘。
容斥做法:
第二类斯特林代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const int maxn = 2e3+5;
ll s[maxn][maxn], c[maxn][maxn], fac[maxn];
void init()
{
memset(s, 0, sizeof(s));
s[1][1] = 1;
for(int i = 2; i < maxn; i++)
for(int j = 1; j <= i; j++)
s[i][j] = (s[i-1][j-1]+(ll)j*s[i-1][j])%mod;
fac[0] = fac[1] = 1;
for(int i = 2; i < maxn; i++)
fac[i] = fac[i-1]*i%mod;
for(int i = 1; i < maxn; i++)
for(int j = 0; j <= i; j++)
{
if(i == j || !j) c[i][j] = 1;
else c[i][j] = (c[i-1][j]+c[i-1][j-1])%mod;
}
}
ll qpow(ll p, int q)
{
ll ans = 1;
while(q)
{
if(q%2) ans = ans*p%mod;
p = p*p%mod;
q /= 2;
}
return ans;
}
int main(void)
{
init();
int n, m, t;
cin >> t;
while(t--)
{
scanf("%d%d", &n, &m);
ll ans = 0;
int e = min(n, m-1);
for(int i = 1; i <= e; i++)
{
ans = (ans + s
[i]*fac[i]%mod*qpow(m-i, n)%mod*c[m][i]%mod)%mod;
}
printf("%lld\n", ans%mod);
}
return 0;
}
容斥代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e3+5;
const int mod = 1e9+7;
typedef long long ll;
ll c[maxn][maxn], rec[maxn], fac[maxn];
void init()
{
fac[0] = fac[1] = 1;
for(int i = 2; i < maxn; i++)
fac[i] = fac[i-1]*i%mod;
for(int i = 1; i < maxn; i++)
for(int j = 0; j <= i; j++)
if(!j || i==j) c[i][j] = 1;
else c[i][j] = (c[i-1][j]+c[i-1][j-1])%mod;
}
ll qmod(ll x, int p)
{
ll ans = 1;
while(p)
{
if(p%2) ans = ans*x%mod;
x = x*x%mod;
p /= 2;
}
return ans;
}
int main(void)
{
init();
int t, n, m;
cin >> t;
while(t--)
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++)
{
rec[i] = qmod(i, n);
for(int j = 1; j < i; j++)
rec[i] = (rec[i]-c[i][j]*rec[j]%mod+mod)%mod;
}
ll ans = 0;
for(int i = 1; i <= min(n, m-1); i++)
ans = (ans+c[m][i]*rec[i]%mod*qmod(m-i, n)%mod)%mod;
printf("%lld\n", ans);
}
return 0;
}
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