题目链接：
题目概述：
计算两个大整数（A,B）的最大公因数
数据范围
0 < A , B ≤ 10 ^
其一在于辗转相减法——辗转相除法的优化（针对大数，避免了大数的模运算带来的多方面的复杂度）
思想就是 以数次 A-B 代替 A%B （这二者是等价的）
辗转相减法：
1 /*
2 Write(X) 输出X
3 Down(X) X除以2
4 Up(X) X乘以2
5 */
6 void Solve(){
7 int t=0;
8 int ok=0;
9 while(A!=B){
10 if(A<B) swap(A,B);
11 int x=A.number[1]%2,y=B.number[1]%2;
12 if(x==0 && y==0){
13 t++;
14 Down(A),Down(B);
15 }
16 else if(x==0) Down(A);
17 else if(y==0) Down(B);
18 else A=A-B;
19 }
20 while(t--) Up(A);
21 Write(A);
22 return ;
23 }
其二就在于压位高精度本身的实现了，我这里选用的是压四位（例如Number[1]=0023，Number[2]=0034表示340023）
结构体（包括 小于比较；不等于比较；减法）
1 #include<map>
2 #include<queue>
3 #include<cmath>
4 #include<cstdio>
5 #include<cstring>
6 #include<iostream>
7 #include<algorithm>
8 using namespace std;
9 typedef long long ll;
10 const int Hash=100003;
11 int A,B,N;
12 int Tot=0;
13 int F[Hash+5];
14 struct data{
15 ll to;
16 int next;
17 ll cost;
18 }E[Hash+5];
19 void Addl(int x,ll y,ll z){
20 // printf("%d %d %d\n",x,y,z);
21 E[++Tot]=(data){y,F[x],z};
22 F[x]=Tot;
23 return ;
24 }
25 int H(ll x){
26 return abs(x%Hash);
27 }
28 int Updata(ll x,int y,ll z){
29 int t=H(x);
30 int i=F[t];
31 while(i){
32 ll c=E[i].to;
33 if(c==x) return E[i].cost;
34 i=E[i].next;
35 }
36 if(y==0) return -1;
37 if(y==1) Addl(t,x,z);
38 return 1;
39 }
40 ll Exgcd(ll a,ll b,ll &x,ll &y){
41 if(b==0){
42 x=1;
43 y=0;
44 return a;
45 }
46 ll res=Exgcd(b,a%b,y,x);
47 y-=a/b*x;
48 return res;
49 }
50 //map<ll,int>mp;
51 void Solve(ll a,ll b,ll c){
52 Tot=0;
53 memset(F,0,sizeof(F));
54 a%=c;
55 ll t=1;
56 ll tmp=1;
57 ll m=sqrt(c);
58 if(m*m<N) m++;
59 Updata(1,1,m+1);
60 // mp.clear();
61 // mp[1]=m+1;
62 for(ll i=1;i<m;i++){
63 tmp=(tmp*a)%c;
64 Updata(tmp,1,i);
65 //if(!mp[tmp]) mp[tmp]=i;
66 }
67 tmp=tmp*a%c;
68 ll d=1,x,y;
69 for(ll i=0;i<m;i++){
70 ll x,y;
71 ll res=Exgcd(d,c,x,y);
72 ll need=(b/res*x%c+c)%(c/res);
73 int t=Updata(need,0,-1);
74 // printf("%d %d<\n",need,t);
75 if(t>0){
76 if(t==m+1) t=0;
77 cout<<i*m+t<<endl;
78 return ;
79 }
80 d=(d*tmp)%c;
81 }
82 printf("-1\n");
83 return ;
84 }
85 int main(){
86 while(scanf("%d%d%d",&A,&B,&N)!=EOF)
87 Solve((ll)A,(ll)B,(ll)N);
88 return 0;
89 }
乘2及除2
1 void Up(alpha &x){
2 int now=1;
3 while(now<=x.large){
4 x.number[now]*=2;
5 now++;
6 }
7
8 now=1;
9 while(now<=x.large){
10 if(x.number[now]>=10000) x.number[now]-=10000,x.number[now+1]++;
11 now++;
12 }
13
14 while(x.number[x.large+1]) x.large++;
15 return ;
16 }
17 void Down(alpha &x){
18 int now=x.large;
19 while(now){
20 if(x.number[now]%2) x.number[now-1]+=10000;
21 x.number[now--]/=2;
22 }
23 while(x.large>1 && !x.number[x.large]) x.large--;
24 return ;
25 }
读入及输出
1 void Read(alpha &x){
2 alpha a;
3 char c=getchar();
4 while(!isdigit(c)) c=getchar();
5 while(isdigit(c)) a.number[++a.large]=c-‘0‘,c=getchar();
6 x.large=(a.large-1)/4+1;
7 reverse(a.number+1,a.number+1+a.large);
8 while(a.large>1 && !a.number[a.large]) a.large--;
9 while(a.large) x.number[(a.large-1)/4+1]=x.number[(a.large-1)/4+1]*10+a.number[a.large--];
10 return ;
11 }
12 void Write(alpha x){
13 alpha a;
14 int front=1;
15 a.large=4*x.large;
16 while(front<=x.large){
17 for(int i=1;i<=4;i++)
18 a.number[(front-1)*4+i]=x.number[front]%10,x.number[front]/=10;
19 front++;
20 }
21 while(a.large>1 && !a.number[a.large]) a.large--;
22 while(a.large) putchar(a.number[a.large--]+‘0‘);
23 putchar(‘\n‘);
24 return ;
25 }
全code
1 /*
2 SuperGcd
3 LG 1414
4 */
5 #include<queue>
6 #include<cstdio>
7 #include<cstring>
8 #include<iostream>
9 #include<algorithm>
10 using namespace std;
11 struct alpha{
12 int large;
13 int number[10005];
14 alpha(){
15 large=0;
16 memset(number,0,sizeof(number));
17 }
18 friend bool operator<(alpha a,alpha b){
19 if(a.large!=b.large) return a.large<b.large;
20 while(a.large)
21 if(a.number[a.large]!=b.number[a.large--]) return a.number[++a.large]<b.number[a.large];
22 return 0;
23 }
24 friend bool operator!=(alpha a,alpha b){
25 if(a.large!=b.large) return 1;
26 while(a.large)
27 if(a.number[a.large]!=b.number[a.large--]) return 1;
28 return 0;
29 }
30 friend alpha operator-(alpha a,alpha b){
31 alpha c;
32 while(c.large<a.large){
33 c.number[++c.large]+=a.number[c.large]-b.number[c.large];
34 while(c.number[c.large]<0) c.number[c.large]+=10000,c.number[c.large+1]--;
35 }
36 while(c.large>1 && !c.number[c.large]) c.large--;
37 return c;
38 }
39 };
40 void Read(alpha &x){
41 alpha a;
42 char c=getchar();
43 while(!isdigit(c)) c=getchar();
44 while(isdigit(c)) a.number[++a.large]=c-‘0‘,c=getchar();
45 x.large=(a.large-1)/4+1;
46 reverse(a.number+1,a.number+1+a.large);
47 while(a.large>1 && !a.number[a.large]) a.large--;
48 while(a.large) x.number[(a.large-1)/4+1]=x.number[(a.large-1)/4+1]*10+a.number[a.large--];
49 return ;
50 }
51 void Write(alpha x){
52 alpha a;
53 int front=1;
54 a.large=4*x.large;
55 while(front<=x.large){
56 for(int i=1;i<=4;i++)
57 a.number[(front-1)*4+i]=x.number[front]%10,x.number[front]/=10;
58 front++;
59 }
60 while(a.large>1 && !a.number[a.large]) a.large--;
61 while(a.large) putchar(a.number[a.large--]+‘0‘);
62 putchar(‘\n‘);
63 return ;
64 }
65 alpha A,B;
66 void Up(alpha &x){
67 int now=1;
68 while(now<=x.large){
69 x.number[now]*=2;
70 now++;
71 }
72
73 now=1;
74 while(now<=x.large){
75 if(x.number[now]>=10000) x.number[now]-=10000,x.number[now+1]++;
76 now++;
77 }
78
79 while(x.number[x.large+1]) x.large++;
80 return ;
81 }
82 void Down(alpha &x){
83 int now=x.large;
84 while(now){
85 if(x.number[now]%2) x.number[now-1]+=10000;
86 x.number[now--]/=2;
87 }
88 while(x.large>1 && !x.number[x.large]) x.large--;
89 return ;
90 }
91 void Solve(){
92 int t=0;
93 int ok=0;
94 while(A!=B){
95 if(A<B) swap(A,B);
96 int x=A.number[1]%2,y=B.number[1]%2;
97 if(x==0 && y==0){
98 t++;
99 Down(A),Down(B);
100 }
101 else if(x==0) Down(A);
102 else if(y==0) Down(B);
103 else A=A-B;
104 }
105 while(t--) Up(A);
106 Write(A);
107 return ;
108 }
109 int main(){
110 Read(A);Read(B);
111 Solve();
112 return 0;
113 }
对于我这样的蒟蒻来讲，这代码可真是难码。。。。。（不过还是很有成就感就是了）
原文地址：https://www.cnblogs.com/Aloyd/p/9296506.html
时间： 2024-11-03 22:13:10