JiuDuOj——1051

#include <iostream>
#include <math.h>
#include <algorithm>
#include <string.h>
 
using namespace std;
 
int main()
{
    int a,n,tmp,k,ans[101];
    while (cin>>a>>n)
    {
        if (a == 0)
        {
            cout<<0;
        }
        else
        {
            memset(ans, 0, sizeof(ans));
            k = 0;
            for (int i = 0; i != n; ++ i)
            {
                tmp = (n - i) * a;
                ans[i] += tmp;
                ans[i + 1] += ans[i] / 10;
                ans[i] = ans[i] % 10;
            }
            for (int i = 0; i != 101; ++ i)
            {
                if (ans[i] != 0)
                {
                    k = i;
                }
            }
            for (int i = k; i != -1; -- i)
            {
                cout<<ans[i];
            }
        }
        cout<<endl;
    }
}

这题...有点卡壳了，特殊情况必须要考虑清楚，还有就是计算的方法需要改进的地方要改进。这题的思路其实，要在草稿纸上把加法竖式列出来就发现，高位其实是n个a相加，往下走最后变成1个a，这样的话就可以归纳成大数加法了。Great

转载于:https://www.cnblogs.com/thewaytomakemiracle/p/5006968.html