ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#206400 | #1745. 一元三次方程求解 | wyz_ | 100 | 1ms | 1244kb | C++11 | 324b | 2024-07-22 17:56:56 | 2024-07-22 20:03:13 |
answer
#include<bits/stdc++.h>
using namespace std;
int cnt;
double a,b,c,d;
int main(){
scanf("%lf %lf %lf %lf",&a,&b,&c,&d);
for(double i = -100; i <= 100; i += 0.01){
if(abs(a*i*i*i + b*i*i + c*i + d) <= 1e-2){
printf("%.2lf ",i);
if(++cnt >= 3)
return 0;
i += 0.99;
}
}
return 0;
}
详细
小提示:点击横条可展开更详细的信息
Test #1:
score: 25
Accepted
time: 0ms
memory: 1236kb
input:
1 -2 -1 2
output:
-1.00 1.00 2.00
result:
ok single line: '-1.00 1.00 2.00 '
Test #2:
score: 25
Accepted
time: 0ms
memory: 1244kb
input:
1 -4.65 2.25 1.4
output:
-0.35 1.00 4.00
result:
ok single line: '-0.35 1.00 4.00 '
Test #3:
score: 25
Accepted
time: 1ms
memory: 1244kb
input:
1 10 -1 -10
output:
-10.00 -1.00 1.00
result:
ok single line: '-10.00 -1.00 1.00 '
Test #4:
score: 25
Accepted
time: 0ms
memory: 1240kb
input:
1 -1.8 -8.59 -0.84
output:
-2.10 -0.10 4.00
result:
ok single line: '-2.10 -0.10 4.00 '