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 |
---|---|---|---|---|---|---|---|---|---|
#206422 | #1745. 一元三次方程求解 | cql | 100 | 0ms | 1288kb | C++11 | 359b | 2024-07-22 18:27:55 | 2024-07-22 20:07:11 |
answer
#include <bits/stdc++.h>
#define int long long
using namespace std;
signed main() {
double a, b, c, d; cin >> a >> b >> c >> d;
for (int j = -10000, cnt = 0; j <= 10000 && cnt < 3; j++) {
double i = (double)j / 100.0;
if (abs(a * i * i * i + b * i * i + c * i + d) <= 0.001) {
printf("%.2f ", i);
j += 99; cnt++;
}
}
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: 1280kb
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: 1288kb
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: 0ms
memory: 1284kb
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: 1284kb
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 '