简单C语言程序计算 求帮忙
后面跟下每个步骤运算结果#includeintmain(){intu[4],a,b,c,x,y,z;scanf("%d%d%d%d",&(u[0]),&(u[1]),&(...
后面跟下每个步骤运算结果
#include
int main(){
int u[4], a, b, c, x, y, z;
scanf("%d %d %d %d", &(u[0]), &(u[1]), &(u[2]), &(u[3]));
a = u[0] + u[1] + u[2] + u[3] - 5;
b = u[0] * (u[1] - u[2] / u[3] + 8);
c = u[0] * u[1] / u[2] * u[3];
x = (a + b + 2) * 3 - u[(c + 3) % 4];
y = (c * 100 - 13) / a / (u[b % 3] * 5);
if ((x + y) % 2 == 0) z = (a + b + c + x + y) / 2;
z = (a + b + c – x - y) * 2;
printf("%d\n", x + y - z);
return 0;
}
输入: 2 5 7 4
输出:( )
算出来的麻烦跟下过程好么? 展开
#include
int main(){
int u[4], a, b, c, x, y, z;
scanf("%d %d %d %d", &(u[0]), &(u[1]), &(u[2]), &(u[3]));
a = u[0] + u[1] + u[2] + u[3] - 5;
b = u[0] * (u[1] - u[2] / u[3] + 8);
c = u[0] * u[1] / u[2] * u[3];
x = (a + b + 2) * 3 - u[(c + 3) % 4];
y = (c * 100 - 13) / a / (u[b % 3] * 5);
if ((x + y) % 2 == 0) z = (a + b + c + x + y) / 2;
z = (a + b + c – x - y) * 2;
printf("%d\n", x + y - z);
return 0;
}
输入: 2 5 7 4
输出:( )
算出来的麻烦跟下过程好么? 展开
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