[PTA]练习8-2 计算两数的和与差
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[PTA]练习8-2 计算两数的和与差
本题要求实现一个计算输入的两数的和与差的简单函数。
函数接口定义:
void sum_diff( float op1, float op2, float *psum, float *pdiff );
其中op1和op2是输入的两个实数,psum和pdiff是计算得出的和与差。
裁判测试程序样例:
#include <stdio.h>
void sum_diff( float op1, float op2, float *psum, float *pdiff );
int main()
{
float a, b, sum, diff;
scanf("%f %f", &a, &b);
sum_diff(a, b, &sum, &diff);
printf("The sum is %.2f\\nThe diff is %.2f\\n", sum, diff);
return 0;
}
/* 你的代码将被嵌在这里 */
输入样例:
4 6
输出样例:
The sum is 10.00
The diff is -2.00
- 提交结果:
- 源码:
#include <stdio.h>
void sum_diff(float op1, float op2, float* psum, float* pdiff);
int main()
{
float a, b, sum, diff;
scanf("%f %f", &a, &b);
sum_diff(a, b, &sum, &diff);
printf("The sum is %.2f\\nThe diff is %.2f\\n", sum, diff);
return 0;
}
/* 你的代码将被嵌在这里 */
void sum_diff(float op1, float op2, float* psum, float* pdiff)
{
*psum = op1 + op2;
*pdiff = op1 - op2;
}
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