3 条题解
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0
#include<stdio.h> #include<math.h> int main() { int x1,y1,x2,y2,x3,y3; double s; scanf("%d %d %d %d %d %d",&x1,&y1,&x2,&y2,&x3,&y3); while(x1!=0||y1!=0||x2!=0||y2!=0||x3!=0||y3!=0) { s=0.5abs((x1-x3)(y2-y3)-(x2-x3)*(y1-y3));
printf("%.1lf\n",s); scanf("%d %d %d %d %d %d",&x1,&y1,&x2,&y2,&x3,&y3); } return 0; }
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#include<stdio.h> #include<math.h> int main() { int x1, y1, x2, y2, x3, y3; double p, p1, p2, p3, s; scanf("%d %d %d %d %d %d", &x1, &y1, &x2, &y2, &x3, &y3); while (x1!= 0||y1!= 0||x2!= 0||y2!= 0||x3!= 0||y3!= 0) { p1 = sqrt(pow(y2 - y1, 2) + pow(x2 - x1, 2)); p2 = sqrt(pow(y3 - y2, 2) + pow(x3 - x2, 2)); p3 = sqrt(pow(y1 - y3, 2) + pow(x1 - x3, 2)); p = (p1 + p2 + p3) /2; s = sqrt(p*(p - p1)*(p - p2)*(p - p3)); printf("%.1f\n", s); scanf("%d %d %d %d %d %d", &x1, &y1, &x2, &y2, &x3, &y3); } return 0; } //float貌似过不去
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信息
- ID
- 146
- 时间
- 3000ms
- 内存
- 128MiB
- 难度
- 6
- 标签
- (无)
- 递交数
- 796
- 已通过
- 245
- 上传者