c语言求利润的编程 [编程入门]利润计算-题解（C语言代码）

```c

#include int jiangjin_1(int);//利润 <= 100000

int jiangjin_2(int);//利润 > 100000 && 利润 <= 200000

int jiangjin_3(int);//利润 > 200000 && 利润 <= 400000

int jiangjin_4(int);//利润 > 400000 && 利润 <= 600000

int jiangjin_5(int);//利润 > 600000 && 利润 <= 1000000

int jiangjin_6(int);//利润 > 1000000

int main(void)

{

int lirun;//利润

scanf("%d", &lirun);

if (lirun <= 100000)

{

printf("%d", jiangjin_1(lirun));

}

else if (lirun > 100000 && lirun <= 200000)

{

printf("%d", jiangjin_2(lirun));

}

else if (lirun > 200000 && lirun <= 400000)

{

printf("%d", jiangjin_3(lirun));

}

else if (lirun > 400000 && lirun <= 600000)

{

printf("%d", jiangjin_4(lirun));

}

else if (lirun > 600000 && lirun <= 1000000)

{

printf("%d", jiangjin_5(lirun));

}

else

{

printf("%d", jiangjin_6(lirun));

}

return 0;

}

int jiangjin_1(int lirun)

{

return lirun * 0.1;

}

int jiangjin_2(int lirun)

{

return 100000 * 0.1 +(lirun - 100000) * 0.075;

}

int jiangjin_3(int lirun)

{

return 100000 * 0.1 + 100000 * 0.075 + (lirun - 200000) * 0.05;

}

int jiangjin_4(int lirun)

{

return 100000 * 0.1 +100000 * 0.075 +200000 * 0.05 +(lirun - 400000) * 0.03;

}

int jiangjin_5(int lirun)

{

return 100000 * 0.1 +100000 * 0.075 +200000 * 0.05 +200000 * 0.03 +(lirun - 600000) * 0.015;

}

int jiangjin_6(int lirun)

{

return 100000 * 0.1 +100000 * 0.075 +200000 * 0.05 +200000 * 0.03 +400000 * 0.015 +(lirun - 1000000) * 0.01;

}

```