UVA 10341 solve it-白红宇

UVA 10341 solve it

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发布时间：2019-06-08

本文共 1747 字，大约阅读时间需要 5 分钟。

Solve the equation:

p∗e−x + q∗sin(x) + r∗cos(x) + s∗tan(x) + t∗x2 + u = 0

where 0 ≤ x ≤ 1.

Input

Input consists of multiple test cases and terminated by an EOF. Each test case consists of 6 integers in a single line: p, q, r, s, t and u (where 0 ≤ p, r ≤ 20 and −20 ≤ q,s,t ≤ 0). There will be maximum 2100 lines in the input ﬁle.

Output

For each set of input, there should be a line containing the value of x, correct up to 4 decimal places, or the string ‘No solution’, whichever is applicable.

Sample Input

0 0 0 0 -2 1 1 0 0 0 -1 2 1 -1 1 -1 -1 1

Sample Output

0.7071 No solution 0.7554

观察题目中的条件，方程必定在定义域内单调递减，逻辑方面就很容易了。

在实现上，使用二分找到一个满足条件的解，不过我一开始使用的whlie循环居然会超时，后来换成for循环就过了，后来才发现我的循环有问题

while(abs(f(mid))>=MIN)            {                if(f(mid)>=MIN)                fir=mid,mid=(fir+las)/2;                else                las=mid,mid=(fir+las)/2;        }

判断结束的精确度不应该是用函数值，而应该是用mid，下面for循环通过代码

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           using namespace std;const int MAX=1e6;const double MIN=1e-6;#define INF 0x7fffffff#define ll long long#define FOR(i,n) for(i=1;i<=n;i++)#define mst(a) memset(a,0,sizeof(a))#define mstn(a,n) memset(a,n,sizeof(a))//struct num{int a,b,i;}a[1005];//bool cmp(const num &x, const num &y){return x.a>y.a;}int p,q,r,s,t,u;float f(float x){ return p*exp(-x)+q*sin(x)+r*cos(x)+s*tan(x)+t*x*x+u;}int main(){ while(scanf("%d%d%d%d%d%d",&p,&q,&r,&s,&t,&u)!=EOF) { double fir=0,las=1,mid=(fir+las)/2; if(f(fir)<0||f(las)>0) cout<<"No solution"<
            
             MIN) fir=mid; else las=mid; } printf("%.4lf\n",mid); 　} 　　} return 0;}

转载于:https://www.cnblogs.com/qq936584671/p/7126179.html

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