快速幂的大数运算_快速幂模

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const long long int mod = 1000000000007;   //对答案取模
int main()
{ 

long long int n;
long long int ans = 1;
long long int temp = 2;
cin >> n;   //求2的n次方
printf("2的%lld次幂对对1000000000007取模的最终值是：", n);
while (n > 0)  //快速幂模板
{ 

if (n%2 == 1)
ans = (ans%mod * temp%mod) % mod;
n /= 2;
temp = (temp % mod * temp % mod)%mod;
}
printf("%lld",ans%mod);
return 0;
}

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
long long int ans[100000];  //存放答案
long long int temp[100000];  //存放2的2^n的值
long long int save[100000];    //临时数组
int ans_len = 1;
int temp_len = 1;
void count_1(long long int* ans, long long int* temp)  //计算数组ans*数组temp,实际上就是简单的高精度大数相乘了
{ 

memset(save, 0, sizeof(save));
for (int i = 0; i < ans_len; i++)  //先不考虑进制
{ 

for (int j = 0; j < temp_len; j++)
{ 

save[i + j] += ans[i] * temp[j];
}
}
int weishu_max = ans_len + temp_len;   //最终答案最终这么多位数。可以那9999*9999这种极端证明
for (int i = 0; i < weishu_max; i++)
{ 

save[i + 1] += save[i] / 10;
save[i] %= 10;
}
if (save[weishu_max - 1] != 0)  //n位数*m位数要么就是m+n位数，要么就是m+n-1位数
ans_len = weishu_max;
else
{ 

ans_len = weishu_max - 1;
}
for (int i = 0; i < ans_len; i++)  //将save的复制给ans
ans[i] = save[i];
}
void count_2(long long int* temp)   //算2的2^n次方,也就是temp自乘
{ 

memset(save, 0, sizeof(save));
for (int i = 0; i < temp_len; i++)
for (int j = 0; j < temp_len; j++)
save[i + j] += temp[i] * temp[j];
int weishu_max = temp_len + temp_len;
for (int i = 0; i < weishu_max; i++)
{ 

save[i + 1] += save[i] / 10;
save[i] %= 10;
}
if (save[weishu_max - 1] != 0)
temp_len = weishu_max;
else
temp_len = weishu_max - 1;
for (int i = 0; i < temp_len; i++)
temp[i] = save[i];
}
int main()
{ 

int n;
cin >> n;
ans[0] = 1;
temp[0] = 2;
while (n > 0)
{ 

if (n % 2 == 1)
{ 

count_1(ans, temp);
}
n /= 2;
count_2(temp);
}
for (int i = ans_len-1; i >=0; i--)
{ 

cout << ans[i];
}
return 0;
}