类欧几里得模板 p5170
1 //类欧几里得的模板题 p5170 2 //求这三个式子； 3 //第一个跟后两个没关联 4 //后两个跟其余两个都有关联； 5 6 7 #include<cstdio> 8 #include<algorithm> 9 #include<math.h> 10 #include<string.h> 11 using namespace std; 12 typedef long long ll; 13 const ll inv2=499122177; 14 const ll inv6=166374059; 15 const ll mod=998244353; 16 int t; 17 ll n,a,b,c; 18 struct query 19 { 20 ll f; 21 ll g; 22 ll h; 23 }; 24 query solve(ll a,ll b,ll c,ll n) 25 { 26 query ans,prec; 27 if(a==0){ 28 ans.f=(b/c)*(n+1)%mod; 29 ans.g=(b/c)*n%mod*(n+1)%mod*inv2%mod; 30 ans.h=(b/c)*(b/c)%mod*(n+1)%mod; 31 } 32 else if(a>=c||b>=c){ 33 prec=solve(a%c,b%c,c,n); 34 ans.f=(prec.f+n*(n+1)%mod*inv2%mod*(a/c)%mod+(n+1)*(b/c)%mod)%mod; 35 ans.g=((a/c)*n%mod*(n+1)%mod*(2*n+1)%mod*inv6%mod+ 36 (b/c)*n%mod*(n+1)%mod*inv2%mod+prec.g)%mod; 37 ans.h=(prec.h+(a/c)*(a/c)%mod*n%mod*(n+1)%mod*(2*n+1)%mod*inv6%mod+ 38 (n+1)*(b/c)%mod*(b/c)%mod+2*(a/c)%mod*prec.g%mod+ 39 2*(b/c)%mod*prec.f%mod+2*(a/c)%mod*(b/c)%mod*n%mod*(n+1)%mod*inv2%mod)%mod; 40 } 41 else{ 42 ll m=(a*n+b)/c; 43 prec=solve(c,c-b-1,a,m-1); 44 ans.f=(n*(m%mod)%mod-prec.f)%mod; 45 ans.g=(n*(n+1)%mod*(m%mod)%mod-prec.f-prec.h)%mod*inv2%mod; 46 ans.h =(n*(m%mod)%mod*((m+1)%mod)%mod-2*prec.g-2*prec.f-ans.f)%mod; 47 } 48 return ans; 49 } 50 int main() 51 { 52 scanf("%d",&t); 53 while(t--){ 54 scanf("%lld%lld%lld%lld",&n,&a,&b,&c); 55 query ans=solve(a,b,c,n); 56 printf("%lld %lld %lld\n", (ans.f + mod) % mod, (ans.h + mod) % mod, (ans.g + mod) % mod); 57 } 58 return 0; 59 }
求第一个式子的模板
1 #include<cstdio> 2 #include<algorithm> 3 #include<math.h> 4 #include<string.h> 5 using namespace std; 6 typedef long long ll; 7 const ll inv2=499122177; 8 const ll inv6=166374059; 9 const ll mod=998244353; 10 int t; 11 ll n,a,b,c; 12 ll solve(ll a,ll b,ll c,ll n) 13 { 14 ll ans,prec; 15 if(a==0) ans=(b/c)*(n+1)%mod; 16 else if(a>=c||b>=c){ 17 prec=solve(a%c,b%c,c,n); 18 ans=(prec+n*(n+1)%mod*inv2%mod*(a/c)%mod+(n+1)*(b/c)%mod)%mod; 19 } 20 else{ 21 ll m=(a*n+b)/c; 22 prec=solve(c,c-b-1,a,m-1); 23 ans=(n*(m%mod)%mod-prec)%mod; 24 } 25 return ans; 26 } 27 int main() 28 { 29 scanf("%d",&t); 30 while(t--){ 31 scanf("%lld%lld%lld%lld",&n,&a,&b,&c); 32 ll ans=solve(a,b,c,n); 33 printf("%lld",(ans+mod)%mod); 34 } 35 return 0; 36 }
