C语言填空:数学分段函数 switch case理解
/*有一函数: y= x (x<10) 输入x的值,求y的值。 y=3x -2 (10≤x<50) y=4x+1 (50≤x<100) y=5x (x≥100)*/ 【6】 void main() { int x,y; 【1】 t; printf("input x=:"); scanf("【2】",&x); if(x<10) 【3】; else if(x>=100) t=10; else 【4】; switch(t) { case 0: y=x; break; case 1: case 2: case 3: case 4: y=3*x-2;break; case 5: case 6: case 7: case 8: case 9: 【5】break; case 10: y=5*x; } printf("y=%d",y); }
/*有一函数: y= x (x<10) 输入x的值,求y的值。 y=3x -2 (10≤x<50) y=4x+1 (50≤x<100) y=5x (x≥100)*/ #include <stdio.h> void main() { int x,y; int t; printf("input x=:"); scanf("%d",&x); if(x<10) t=0; else if(x>=100) t=10; else t=x/10; switch(t) { case 0: y=x; break; case 1: case 2: case 3: case 4: y=3*x-2;break; case 5: case 6: case 7: case 8: case 9: y=4*x+1;break; case 10: y=5*x; } printf("y=%d",y); }