【BZOJ】【2242】【SDOI2011】计算器
快速幂/扩展欧几里得/BSGS
经典好例题！！
三个问题三种算法……
算法：白书（算法竞赛入门经典——训练指南）全有……
1 /************************************************************** 2 Problem: 2242 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:1824 ms 7 Memory:2476 kb 8 ****************************************************************/ 9 10 //BZOJ 2242 11 #include<map> 12 #include<cmath> 13 #include<cstdio> 14 #include<cstdlib> 15 #include<cstring> 16 #include<iostream> 17 #include<algorithm> 18 #define rep(i,n) for(int i=0;i<n;++i) 19 #define F(i,j,n) for(int i=j;i<=n;++i) 20 #define D(i,j,n) for(int i=j;i>=n;--i) 21 using namespace std; 22 23 int getint(){ 24 int v=0,sign=1; char ch=getchar(); 25 while(ch<'0'||ch>'9') {if (ch=='-') sign=-1; ch=getchar();} 26 while(ch>='0'&&ch<='9') {v=v*10+ch-'0'; ch=getchar();} 27 return v*=sign; 28 } 29 /*******************tamplate********************/ 30 typedef long long LL; 31 LL x,y,z,P; 32 LL pow(LL a,LL b,LL P){ 33 LL r=1,base=a; 34 while(b){ 35 if (b&1) r=r*base%P; 36 base=base*base%P; 37 b>>=1; 38 } 39 return r; 40 } 41 void exgcd(LL a,LL b,LL &d,LL &x,LL &y){ 42 if (!b) {d=a; x=1; y=0; return;} 43 else{exgcd(b,a%b,d,y,x); y-=x*(a/b);} 44 } 45 LL log_mod(LL a,LL b,LL P){ 46 LL m,v,e=1,i; 47 m=ceil(sqrt(P+0.5));//这里需要用ceil……？ 48 v=pow(a,P-m-1,P);//inv 49 map<LL,LL> x; 50 x[1]=0; 51 for(int i=1;i<m;++i){ 52 e=e*a%P; 53 if (!x.count(e)) x[e]=i; 54 } 55 rep(i,m){ 56 if (x.count(b)) return (LL)i*m+x[b]; 57 b=b*v%P; 58 } 59 return -1; 60 } 61 int main(){ 62 int T=getint(),K=getint(); 63 while(T--){ 64 y=getint(); z=getint(); P=getint(); 65 if (K==1) printf("%lld\n",pow(y,z,P)); 66 if (K==2){ 67 LL d=0,X=0,Y=0; 68 exgcd(y,P,d,X,Y); 69 if(z%d){ printf("Orz, I cannot find x!\n"); continue; } 70 X*=z/d; 71 LL T=X*d/P; 72 X-=T*P/d; 73 if (X<0) X+=P/d; 74 printf("%lld\n",X); 75 } 76 if (K==3){ 77 y%=P; z%=P; 78 if(!y && !z) {printf("1\n"); continue;} 79 if(!y) { printf("Orz, I cannot find x!\n"); continue; } 80 LL ans=log_mod(y,z,P); 81 if (ans==-1){ printf("Orz, I cannot find x!\n"); continue; } 82 printf("%lld\n",ans); 83 } 84 } 85 return 0; 86 }
