考的是if()
的使用,还要给三条边判断大小
判断优先级:
判断按题给顺序来
#include
int main()
{
int a = 0, b = 0, c = 0, x = 0, y = 0, z = 0;
scanf("%d %d %d", &a, &b, &c);
x = (a < b ? a : b) < c ? (a < b ? a : b) : c;
z = (a > b ? a : b) > c ? (a > b ? a : b) : c;
y = a + b + c - x - z;
if (x + y <= z)
{
printf("Not triangle\n");
}
else
{
if (x * x + y * y == z * z) printf("Right triangle\n");
else if (x * x + y * y > z * z) printf("Acute triangle\n");
else if (x * x + y * y < z * z) printf("Obtuse triangle\n");
if(x == y || y == z || x == z) printf("Isosceles triangle\n");
if(x == y && y == z) printf("Equilateral triangle\n");
}
return 0;
}
数据类型用double
先计算,后判断,超重要多输出个BMI
看着麻烦,刚开始把我也唬住了
其实用%g
就好了,涨知识了(参考群里《C Primer Plus》表4.3)
#include
int main()
{
double m = 0, h = 0, bmi = 0;
scanf("%lf %lf", &m, &h);
bmi = m / (h * h);
if (bmi < 18.5) printf("Underweight\n");
else if (bmi >= 24) printf("%.6g\nOverweight\n", bmi);
else printf("Normal\n");
return 0;
}
对用电量判断,最后保留一位小数
#include
int main()
{
double a = 0, ans = 0;
scanf("%lf", &a);
if (a <= 150) ans = a * 0.4463;
else if (a >= 151 && a <= 400) ans = 150 * 0.4463 + (a - 150) * 0.4663;
else ans = 150 * 0.4463 + 250 * 0.4663 + (a - 400) * 0.5663;
printf("%.1lf", ans);
return 0;
}
先比大小,得出ABC
ABC、ACB、BAC、BCA、CAB、CBA六种情况,不用字符串也可以,就麻烦一点
#include
int main()
{
int a = 0, b = 0, c = 0, A = 0, B = 0, C = 0;
char x = 0, y = 0, z = 0;
scanf("%d %d %d", &a, &b, &c);
scanf("\n%c%c%c", &x, &y, &z);
A = (a < b ? a : b) < c ? (a < b ? a : b) : c;
C = (a > b ? a : b) > c ? (a > b ? a : b) : c;
B = a + b + c - A - C;
if (x == 'A' && y == 'B' && z =='C') printf("%d %d %d", A, B, C);
else if (x == 'A' && y == 'C' && z =='B') printf("%d %d %d", A, C, B);
else if (x == 'B' && y == 'A' && z =='C') printf("%d %d %d", B, A, C);
else if (x == 'B' && y == 'C' && z =='A') printf("%d %d %d", B, C, A);
else if (x == 'C' && y == 'B' && z =='A') printf("%d %d %d", C, B, A);
else if (x == 'C' && y == 'A' && z =='B') printf("%d %d %d", C, A, B);
return 0;
}
#include
int main()
{
int n = 0, n1 = 0, v1 = 0, n2 = 0, v2 = 0, n3 = 0, v3 = 0;
int x = 0, y = 0, z = 0, ans = 0;
scanf("%d %d %d %d %d %d %d", &n, &n1, &v1, &n2, &v2, &n3, &v3);
if (n % n1 == 0) x = (n / n1) * v1;
else x = (n / n1 + 1) * v1;
if (n % n2 == 0) y = (n / n2) * v2;
else y = (n / n2 + 1) * v2;
if (n % n3 == 0) z = (n / n3) * v3;
else z = (n / n3 + 1) * v3;
ans = (x < y ? x : y) < z ? (x < y ? x : y) : z;
printf("%d", ans);
return 0;
}
洛谷题确实有点难度,不过再难也是基础题。加油吧,少年!
也可以试试C语言网的题集,更基础一点
晚安