GCD HDU - 1695

題意：

給出a,b,c,d,k,求出a<=x<=b, c<=y<=d 且gcd(x,y) == k 的（x,y）的對數。
求的是不同數量對的總數

題解：

和這個題一樣P3455 [POI2007]ZAP-Queries，但是本題要求求不同數量對的總數，所以最后的結果要減去重復值
如果是莫比烏斯+分塊做法，對于區間[1,b],[1,d],b<d,重復部分是[1,b]部分，所以減去solve(b,b)/2
如果是容斥做法，對于每個i∈[1,b],求[1,d]中互質的數量，那我們減去i與[1,i]中互質的數量
詳細看代碼

代碼：

莫比烏斯+分塊

#include <bits/stdc++.h> #include <unordered_map> #define debug(a, b) printf("%s = %d\n", a, b); using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> PII; clock_t startTime, endTime; //Fe~Jozky const ll INF_ll= 1e18; const int INF_int= 0x3f3f3f3f; void read(){}; template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar) {x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...); } template <typename T> inline void write(T x) {if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0'); } void rd_test() { #ifdef ONLINE_JUDGE #elsestartTime = clock ();freopen("data.in", "r", stdin); #endif } void Time_test() { #ifdef ONLINE_JUDGE #elseendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC); #endif } const int maxn=2e6+9; int prime[maxn]; int mu[maxn]; int sum[maxn]; int vis[maxn]; int cnt=0; void get_mu(int N){mu[1]=1;vis[1]=vis[0]=1;for(int i=2;i<=N;i++){if(!vis[i]){prime[++cnt]=i;mu[i]=-1;}for(int j=1;j<=cnt&&i*prime[j]<=N;j++){vis[i*prime[j]]=1;if(i%prime[j]==0)break;mu[i*prime[j]]=-mu[i];}}for(int i=1;i<=N;i++){sum[i]=sum[i-1]+mu[i];} } ll solve(int a,int b,int k){a/=k;b/=k;int minn=min(a,b);ll ans=0;for(int l=1,r;l<=minn;l=r+1){r=min(a/(a/l),b/(b/l));ans+=1ll*(sum[r]-sum[l-1])*(a/l)*(b/l);}return ans; } int main() {//rd_test();get_mu(1000000);int t;read(t);int cas=0;while(t--){int a,b,c,d,k;read(a,b,c,d,k);if(k==0){printf("Case %d: 0\n",++cas);continue;}if(b>d)swap(b,d); // printf("%d %d %d %d\n",a,b,c,d); printf("Case %d: %lld\n",++cas,solve(b,d,k)-solve(b,b,k)/2);}//Time_test(); }

容斥

#include <bits/stdc++.h> #include <unordered_map> #define debug(a, b) printf("%s = %d\n", a, b); using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> PII; clock_t startTime, endTime; //Fe~Jozky const ll INF_ll= 1e18; const int INF_int= 0x3f3f3f3f; void read(){}; template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar) {x= 0;char c= getchar();bool flag= 0;while (c < '0' || c > '9')flag|= (c == '-'), c= getchar();while (c >= '0' && c <= '9')x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();if (flag)x= -x;read(Ar...); } template <typename T> inline void write(T x) {if (x < 0) {x= ~(x - 1);putchar('-');}if (x > 9)write(x / 10);putchar(x % 10 + '0'); } void rd_test() { #ifdef ONLINE_JUDGE #elsestartTime = clock ();freopen("data.in", "r", stdin); #endif } void Time_test() { #ifdef ONLINE_JUDGE #elseendTime= clock();printf("\nRun Time:%lfs\n", (double)(endTime - startTime) / CLOCKS_PER_SEC); #endif } const int maxn=3e5+9; int prime[maxn]; int cnt=0; void divide(int n){cnt=0;for(int i=2;i*i<=n;i++){if(n%i==0){prime[cnt++]=i;while(n%i==0)n/=i;}}if(n!=1)prime[cnt++]=n; } int solve(int S){int ans=0;for(int i=1;i<(1<<cnt);i++){int tmp=1;int num=0;for(int j=0;j<cnt;j++){if(i&(1<<j)){tmp*=prime[j];num++;}}if(num&1)ans+=S/tmp;else ans-=S/tmp;}return S-ans; } int main() {//rd_test();int t;read(t);int cas=0;while(t--){int a,b,c,d,k;read(a,b,c,d,k);if(k==0){printf("Case %d: 0\n",++cas);continue;}if(b>d)swap(b,d);b/=k;d/=k;ll ans=0;for(int i=1;i<=b;i++){divide(i);ans+=solve(d)-solve(i-1);}printf("Case %d: %lld\n",++cas,ans);}return 0;//Time_test(); }

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