#include <stdio.h>

int main()

{

? ? int x,y;

? ? x = 10;

? ? y = -3;

? ? int a = x+y;

? ? int b = x-y;

? ? int c = x*y;

? ? int d = x/y;

? ? int e = x%y;

? ? printf("x+y=%d\n",a);??

? ? printf("x-y=%d\n",b);

? ? printf("x*y=%d\n",c);

? ? printf("x/y=%d\n",d);

? ? printf("x%%y=%d\n",e);??

? ? return 0;? ??

}

再給定義五個量表示運(yùn)算之后的結(jié)果，根據(jù)題目賦予這五個量表達(dá)式，最后把這五個量放進(jìn)輸出

#include <stdio.h>

int main()
{
??? int x,y,M1,M2,M3,M4,M5;
??? x = 10;
??? y = -3;
??? M1=x+y;
??? M2=x-y;
??? M3=x*y;
??? M4=x/y;
??? M5=x%y;
??? printf("x+y=%d\n", M1??? );?
??? printf("x-y=%d\n",? M2?? );
??? printf("x*y=%d\n",? M3?? );
??? printf("x/y=%d\n",? M4?? );
??? printf("x%%y=%d\n",? M5?? );?
??? return 0;???
}

上面說的在第六行，直接把第七行復(fù)制粘貼到第六行，上述直接給出x=5，y=8.? ??現(xiàn)在要z = x*8/y+2=4？所以加個括號足矣，?z = x*8/(y+2)