[HDU1695]GCD
[HAOI2011]Problem b
[POI2007]ZAP-Queries
令$$ans(n, m)=sum_{i=1}nsum_{j=1}m[GCD(i, j) == k]$$
[=sum_{i=1}^{lfloorfrac{n}{k}
floor}sum_{j=1}^{lfloorfrac{m}{k}
floor}[GCD(i, j) == 1]
]
令$$f(d)=sum_{i=1}{lfloorfrac{n}{k} floor}sum_{j=1}{lfloorfrac{m}{k} floor}[GCD(i, j) == d]$$
[g(x)=sum_{x|d}f(d)
]
[=sum_{i=1}^{lfloorfrac{n}{k}
floor}sum_{j=1}^{lfloorfrac{m}{k}
floor}[x|GCD(i, j)]
]
[=sum_{i=1}^{lfloorfrac{n}{xk}
floor}sum_{j=1}^{lfloorfrac{m}{xk}
floor}[1|GCD(i, j)]
]
[=lfloorfrac{n}{xk}
floorlfloorfrac{m}{xk}
floor
]
[f(n)=sum_{n|d}mu(frac{d}{n})g(d)
]
则$$ans(n, m)=f(1)=sum_{d=1}^nmu(d)g(d)$$
[=sum_{d=1}^nmu(d)lfloorfrac{n}{dk}
floorlfloorfrac{m}{dk}
floor
]
[HDU1695]GCD
又((x, y))和((y,x))是等价的，要减去重复算的
假设(b<d),则最终(ans=ans(b, d)-ans(b, b)/2)
void init(){
miu[1]=1;
for(int i=2; i < N; i++) {
if(!p[i]) p[++p[0]]=i, miu[i]=-1;
for(int j=1; j <= p[0] && i*p[j] < N; j++){
p[i*p[j]]=1; if(i%p[j] == 0) {miu[i*p[j]]=0; break;} else miu[i*p[j]]=-miu[i];
}
}
}
void solve(){
init(); int T=read();
for(int i=1; i <= T; i++){
int a=read(), b=read(), c=read(), d=read(), k=read(); cout<<"Case "<<i<<": ";
if(k == 0) {cout<<0<<endl; continue;} b/=k, d/=k;
ll ans1=0, ans2=0, mn=min(b, d);
for(int i=1; i <= mn; i++)
ans1+=1LL*miu[i]*(b/i)*(d/i), ans2+=1LL*miu[i]*(mn/i)*(mn/i);
cout<<ans1-ans2/2<<endl;
}
}
[HAOI2011]Problem b
这题要容斥一下(ans=ans(b, d)-ans(a-1, d)-ans(b, c-1)+ans(a-1, c-1)),
还要整除分块，否则会(TLE)
void init(){
p[1]=miu[1]=1;
for(int i=2; i < N; i++) {
if(!p[i]) p[++p[0]]=i, miu[i]=-1;
for(int j=1, x; j <= p[0] && (x=p[j]*i) < N; j++){ p[x]=1;
if(i%p[j] == 0) {miu[x]=0; break;} miu[x]=-miu[i];
}
}
for(int i=1; i < N; i++) miu[i]+=miu[i-1];
}
ll cal(ll m, ll n){ m/=k, n/=k;
ll ans=0, mn=min(n, m);
for(int i=1, j; i <= mn; i=j+1){
j=min(n/(n/i), m/(m/i));
ans+=1LL*(miu[j]-miu[i-1])*(n/i)*(m/i);
}
return ans;
}
void solve(){
int T=read(); init();
while(T--){
ll a=read(), b=read(), c=read(), d=read(); k=read();
printf("%lld
", cal(b, d)-cal(a-1, d)-cal(b, c-1)+cal(a-1, c-1));
}
}
[POI2007]ZAP-Queries 这题(ans=ans(n, m))