In c++ code, given that m=11,n=41, output m and n to the second, third, and fourth powers.

diexinggui 注册会员
2023-02-28 18:20

<< m*m << endl; cout<<"n的2次方: "< ; cout<< "m的3次方: " << m*m *m<< endl; cout<<"n的3次方: "< ; cout<< "m的4次方: " << m*m *m*m<< endl; cout<<"n的4次方: "< ; return 0; } .

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czyroy 注册会员
2023-02-28 18:20

,input[inputid],powinput[powid], pow (input[inputid],powinput[powid])); } } return 0 ; } .

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linyu3094 注册会员
2023-02-28 18:20

< div class = "md_content_show e397 data - v - 3967" = "" >

hope to adopt

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diaotianming1988 注册会员
2023-02-28 18:20

This is an algorithm, so it looks a little bit more, but it's actually good

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c26533806 注册会员
2023-02-28 18:20

,sum); //因为要求每个数据占8列，所以在这里加个\t，为占位空格操作符，为四个空格，这样每打印一个数据空4个字符，两个字符间就有8个了； } } printf ( "\n" ); //因为要求左对齐，所以在两个循环间加个换行操作； sum = 1 ; //这里将sum再次初始化为1 for (i = 1 ; i <= 4 ; i++) //同上，开启计算n次方的循环； { sum = sum * n; if (i == 2 || i == 3 || i == 4 ) { printf ( "%d\t" , sum); } } return 0 ; } .

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In c++ code, given that m=11,n=41, output m and n to the second, third, and fourth powers.

In c++ code, given that m=11,n=41, output m and n to the second, third, and fourth powers.

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