1.大整数加法

题意：计算两个数得和

#include <bits/stdc++.h>
using namespace std;
int main()
{
    char a[105],b[105];
    int  ans[105] = {1};
    memset(ans,0,sizeof(ans));
    scanf("%s",a);
    scanf("%s",b);
    for (int i = 0; i < strlen(a) ; i++)
    {
        ans[strlen(a)-i] += a[i] - ‘0‘;
    }
    for (int i = 0; i < strlen(b) ; i++)
    {
        ans[strlen(b)-i] += b[i] - ‘0‘;
    }
    ans[0] = max(strlen(a),strlen(b));
    for (int i=1; i <=  ans[0] ; i++)
    {
        if ( ans[i] < 10)continue;
        ans[i+1] +=  ans[i] / 10;
        ans[i] %= 10;

        if (i+1 > ans[0])  ans[0] = i+1;
    }
     for (int i=ans[0]; i >=  1 ; i--)
    {
        printf("%d",ans[i]);
    }
    printf("\n");
    return 0;
}

2.大整数乘法

题意：计算2 ^1000 各位数字累加和

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int VAL = 1000;
int32_t main()
{
    char a[N],b[N];
    int  ans[N] ;
    ans[0] = 1;
    ans[1] = 1;
    for (int32_t i = 0; i < 100; i++)
    {
        for (int32_t j=1; j <= ans[0] ; j++)
        {
            ans[j] *= 1024;
        }
        for (int32_t j=1; j <= ans[0]; j++)
        {
            if (ans[j] < VAL) continue;
            ans[j+1] += ans[j] / VAL;
            ans[j] %= VAL;
            if (j == ans[0]) ++ans[0];
        }
    }
    int sum = 0;
    for (int i=1; i <= ans[0]; i++)
    {
        while (ans[i])
        {
            sum += ans[i]%10;
            ans[i] /= 10;
        }
    }
    cout<<sum<<endl;
    return 0;
}

注意：VAL 的设置：是让每位ans[i] 保存三位数，一定程度可以减少数组的长度