请求出错错误代码503请尝试刷新页面重试 1 #include <stdio.h> 2 #include <string.h> 3 4 typedef long long int lli; 5 6 lli power_digui(lli m, lli n) // 递归版 7{ 8 if (n == 0) // 任何数的0次方为1 9 return 1; 10 lli temp = power_digui(m, n >> 1); 11 temp = ...
return 0; } 这段代码包含了两个快速幂算法函数:一个不带取模运算,一个带取模运算。同时,还包含了测试这两个函数的 main 函数。
将上述的代码结构化,也就是写成函数: int PowerMod(int a, int b, int c) { int ans = 1; a = a % c; while(b>0) { if(b % 2 = = 1) ans = (ans * a) % c; b = b/2; a = (a * a) % c; } return ans; } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 1...
x^y可如下实现 unsigned long pow(int x, unsigned y){ unsigned long tmp;if(!y) return 1;tmp = pow(x, y / 2);if(y % 2 == 0) return (tmp * tmp);else return (tmp * tmp * x);}
n,mn,len,lenm,i,mnl:longint;a,m:arrty;procedure mxm;var c:arrty;i,j:longint;begin fillchar(c,sizeof(c),0);for i:=1 to lenm do for j:=1 to lenm do begin inc(c[i+j-1],m[i]*m[j]);inc(c[i+j],c[i+j-1] div 10);c[i+j-1]:=c[i+j-1] mod ...
这个题就是快速幂,注意特判,一开始忘了特判,wa了一发。 代码: 1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<cmath> 5 #include<cstdlib> 6 #include<algorithm> 7 using namespacestd; 8 typedef long longll;
快速幂主要试求a^b mod c，可以百度模取幂运算
18 importjava.util.Scanner; publicclassMain { publicstaticvoidmain(String[] args) { Scanner sc =newScanner(System.in); longa=sc.nextLong(); longb=sc.nextLong(); longm=sc.nextLong(); longres=1%m; while(b>0){ if(b%2==1) res=res*a%m; ...
AC代码 #include<cstdio>#include<cstring>#include<ctype.h>#include<cstdlib>#include<cmath>#include<climits>#include<ctime>#include<iostream>#include<algorithm>#include<deque>#include<vector>#include<queue>#include<string>#include#include<stack>#include<set>#include<numeric>#include<sstream>#include...
数学基础知识(高精、快速幂、龟速乘……),压位高精模板代码龟速乘llch(llx,lly){llret=0;while(y){if(y&1)ret=(ret+x)%m;x=(x+x)%m,y>>=1;}returnret;}...