Can you solve this equation?(二分查找的简单应用)
2013-01-27 11:15
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Can you solve this equation?
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
SubmitStatus
Description
Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.
Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has a real number Y (fabs(Y) <= 1e10);
Output
For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.
Sample Input
Sample Output
思路:
记f(x)为等式左边的值,如果f(100.0)小于Y或f(0.0)大于Y,无解,否则必有解,且解的范围为[0.0, 100.0],不断二分区间,直至找到f(x) = Y
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
SubmitStatus
Description
Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.
Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has a real number Y (fabs(Y) <= 1e10);
Output
For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.
Sample Input
2 100 -4
Sample Output
1.6152 No solution!
思路:
记f(x)为等式左边的值,如果f(100.0)小于Y或f(0.0)大于Y,无解,否则必有解,且解的范围为[0.0, 100.0],不断二分区间,直至找到f(x) = Y
#include <iostream> #include <string> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <map> #include <vector> #include <queue> #include <stack> #define LL long long #define MAXI 2147483647 #define MAXL 9223372036854775807 #define eps (1e-8) #define dg(i) cout << "*" << i << endl; using namespace std; double Solve(double x) { return (8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6); } int main() { int t; double y, low, high, mid; scanf("%d", &t); while(t--) { scanf("%lf",&y); low = 0.0; high = 100.0; if(y > Solve(100.0) || y < Solve(0.0)) puts("No solution!"); else { while(low + eps < high) { mid = (low + high) * 0.5; if(y > Solve(mid) + eps) low = mid; else if(y < Solve(mid) - eps) high = mid; else break; } printf("%.4lf\n", mid); } } return 0; }
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