Zap
FGD正在破解一段密码，他需要回答很多类似的问题：对于给定的整数a,b和d，有多少正整数对x,y，满足x<=a，y<=b，并且gcd(x,y)=d。作为FGD的同学，FGD希望得到你的帮助。
Input
第一行包含一个正整数n，表示一共有n组询问。（1<=n<= 50000）接下来n行，每行表示一个询问，每行三个
正整数，分别为a,b,d。（1<=d<=a,b<=50000）
Output
对于每组询问，输出到输出文件zap.out一个正整数，表示满足条件的整数对数。
Sample Input
2
4 5 2
6 4 3
Sample Output
3
2
//对于第一组询问，满足条件的整数对有(2,2)，（2,4），（4，2）。对于第二组询问，满足条件的整数对有（
6,3），（3,3）。
题解
原问题等价于求有多少对二元组((x,y))满足(xle a/k,yle b/k)且(x,y)互质。
设(D(a,b,k))表示满足(x le a,y le b)且(k|gcd(x,y))的二元组有多少对，显然(D(a,b,k)=lfloor a/k floorlfloor b/k floor)。
设(F(a,b))表示(xle a,yle b)且(x,y)互质的二元组有多少对，由容斥原理可得：
上式表示没有限制的是(D(a,b,1))，减去公因数至少有一个质因子的(D(a,b,p))，加上减重复了的。以此类推，(D(a,b,i))的系数恰好就是Mobius函数。
另外(D(a,b,i))在(forall i in [x,min(lfloor a/lfloor a/x floor floor,lfloor b/lfloor b/x floor floor)])的值都相等，可以整除分块。那么预处理出Mobius函数的前缀和，就可直接累加这一段的答案。时间复杂度(O(N log log N +N( sqrt{a}+sqrt{b})))。
#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
rg T data=0,w=1;
rg char ch=getchar();
while(!isdigit(ch)){
if(ch=='-') w=-1;
ch=getchar();
}
while(isdigit(ch))
data=data*10+ch-'0',ch=getchar();
return data*w;
}
template<class T>il T read(rg T&x){
return x=read<T>();
}
typedef long long ll;
co int N=5e4+1;
int miu[N];
bool v[N];
void Zap(){
int a,b,k,ans=0;
read(a),read(b),read(k);
a/=k,b/=k;
if(a>b) std::swap(a,b);
for(int x=1,gx;x<=a;x=gx+1){
gx=std::min(a/(a/x),b/(b/x));
ans+=(miu[gx]-miu[x-1])*(a/x)*(b/x);
}
printf("%d
",ans);
}
int main(){
// freopen(".in","r",stdin),freopen(".out","w",stdout);
for(int i=1;i<N;++i) miu[i]=1,v[i]=0;
for(int i=2;i<N;++i){
if(v[i]) continue;
miu[i]=-1;
for(int j=2*i;j<N;j+=i){
v[j]=1;
if(j/i%i==0) miu[j]=0;
else miu[j]*=-1;
}
}
for(int i=1;i<N;++i) miu[i]+=miu[i-1];
int n=read<int>();
while(n--) Zap();
return 0;
}
再分析
重温了一遍洛谷网课上AmberFrame的ppt。
Devu and Flowers
Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the same color.
Now Devu wants to select exactly s flowers from the boxes to decorate his garden. Devu would like to know, in how many different ways can he select the flowers from each box? Since this number may be very large, he asks you to find the number modulo (109 + 7).
Devu considers two ways different if there is at least one box from which different number of flowers are selected in these two ways.
Input
The first line of input contains two space-separated integers n and s (1 ≤ n ≤ 20, 0 ≤ s ≤ 1014).
The second line contains n space-separated integers f1, f2, … fn (0 ≤ fi ≤ 1012).
Output
Output a single integer — the number of ways in which Devu can select the flowers modulo (109 + 7).
Sample test(s)
input
2 3
1 3
1
2
2 3
1 3
output
2
1
2
input
2 4
2 2
1
2
2 4
2 2
output
1
1
1
input
3 5
1 3 2
1
2
3 5
1 3 2
output
3
1
3
Note
Sample 1. There are two ways of selecting 3 flowers: {1, 2} and {0, 3}.
Sample 2. There is only one way of selecting 4 flowers: {2, 2}.
Sample 3. There are three ways of selecting 5 flowers: {1, 2, 2}, {0, 3, 2}, and {1, 3, 1}.
分析
多重集组合问题。根据结论，答案为
由于N很小，所以直接枚举子集即可。算组合数的时候用for循环算。时间复杂度(O(2^N N))
#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
rg T data=0,w=1;
rg char ch=getchar();
while(!isdigit(ch)){
if(ch=='-') w=-1;
ch=getchar();
}
while(isdigit(ch))
data=data*10+ch-'0',ch=getchar();
return data*w;
}
template<class T>il T read(rg T&x){
return x=read<T>();
}
typedef long long ll;
co int mod=1e9+7;
ll a[22],m,ans=0;
int inv[22],n;
int power(int a,int b){
int c=1;
for(;b;b>>=1){
if(b&1) c=(ll)c*a%mod;
a=(ll)a*a%mod;
}
return c;
}
int C(ll y,int x){
if(y<0||x<0||y<x) return 0;
y%=mod;
if(y==0||x==0) return 1;
int ans=1;
for(int i=0;i<x;++i)
ans=(ll)ans*(y-i)%mod;
for(int i=1;i<=x;++i)
ans=(ll)ans*inv[i]%mod;
return ans;
}
int main(){
// freopen(".in","r",stdin),freopen(".out","w",stdout);
for(int i=1;i<=20;++i) inv[i]=power(i,mod-2);
read(n),read(m);
for(int i=0;i<n;++i) read(a[i]);
for(int x=0;x<1<<n;++x){
ll t=n+m;
int p=0;
for(int i=0;i<n;++i)
if(x>>i&1) ++p,t-=a[i];
t-=p+1;
if(p&1) ans=(ans-C(t,n-1))%mod;
else ans=(ans+C(t,n-1))%mod;
}
printf("%lld
",(ans+mod)%mod);
return 0;
}