一元三次方程求解

题目分析：

  1. 根的范围在-100到100之间
  2. 根与根之间差的绝对值>=1，我们可以用二分判断答案所在最小区间1，求解答案区间在[i,i + 1] (-100<=i<100)
  3. 实数范围用浮点数二分

代码如下：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<set>

using namespace std;

#define inf 0x3f3f3f3f
#define eps 1e-6
#define db double
#define ll long long
#define  mm(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define pb push_back
#define el endl
#define debug(x) cerr<<#x<<" = "<<x
#define fgx cerr<<"-------------------------"<<endl
#define shutio ios::sync_with_stdio(false),cin.tie(0)
#define  mk make_pair
#define lowbit(x) (x) & (-x)
#define  fi first
#define se second

db a,b,c,d;

db fc(db x) {
    return a * x * x * x + b * x * x + c * x + d;
}

int main() {
    shutio;
    cin >> a >> b >> c >> d;
    int s = 0;
    for(int i = -100; i < 100; i ++ ) {
        db l = i;
        db r = i + 1;
        db fa = fc(l);
        db fb = fc(r);
        if(!fa) {
            printf("%.2f ",l);
            s ++;
            continue;
        }
        if(fa * fb < 0) {
            while(r - l > eps) {
                db mid = (l + r) / 2;
                if(fc(mid) * fc(r) >= 0) r = mid;
                else l = mid;
            }
            s ++;
            printf("%.2f ",l);
        }
        if(s == 3) break;
    }
    return 0;
}