【C++】高精度方法求2的N次方

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2的N次方计算中结果可能超过long的范围

#include <iostream>#include <string>using namespace std;void mult(int m[],int a);void multipy(int n[],int m[]);int mini(int a);int main(void){const int max = 200;int m[max] = {0};m[max-1] = 2;int n[max] = {0};int a;cin >> a;if(a%2==1)n[max-1] = 2;else    n[max-1] = 1;while(a>1){int s = mini(a);for(int i = 0;i<max-1;i++)m[i] = 0;    m[max-1] = 2;    mult(m,s);multipy(n,m);a -= s;}int z = 0;while(n[z]==0)z++;for(int i = z;i<max;i++)cout << n[i];}void mult(int *m,int a){const int max = 200;if(a==2){int y = 0;int e[max] = {0};for(int i = max-1;i>=0;i--){    for(int j = max-1;j>=0;j--)    e[j-y] += m[i]*m[j];y++; }for(int i = max-1;i>0;i--){if(e[i]>9){    e[i-1] += e[i]/10;    e[i] %= 10;    }}for(int i = max-1;i>=0;i--)m[i] = e[i];}else{int y = 0;int e[max] = {0};for(int i = max-1;i>=0;i--){    for(int j = max-1;j>=0;j--)    e[j-y] += m[i]*m[j];y++; }for(int i = max-1;i>0;i--){if(e[i]>9){    e[i-1] += e[i]/10;    e[i] %= 10;    }}for(int i = max-1;i>=0;i--)m[i] = e[i];mult(m,a/2);    }}void multipy(int *n,int *m){const int max = 200;int y = 0;int e[max] = {0};for(int i = max-1;i>=0;i--){    for(int j = max-1;j>=0;j--)    e[j-y] += n[i]*m[j];y++; }for(int i = max-1;i>0;i--){if(e[i]>9){    e[i-1] += e[i]/10;    e[i] %= 10;    }}for(int i = max-1;i>=0;i--)n[i] = e[i];}int mini(int a){int k = 1;while(k<=a)k *= 2;return k/2;}

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