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POJ - 3299 Humidex
题意:已知两数,根据公式求第三个数。
分析:
1、.lfG++编译不过的C++可能编译过。
2、输出.lf改成.f后G++可编译过。
#pragma comment(linker, "/STACK:102400000, 102400000")#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<iomanip>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define Min(a, b) ((a < b) ? a : b)#define Max(a, b) ((a < b) ? b : a)const double eps = 1e-8;inline int dcmp(double a, double b){ if(fabs(a - b) < eps) return 0; return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 10000 + 10;const int MAXT = 10000 + 10;using namespace std;int main(){ char a; while(1){ double T = 200.0, D = 200.0, H = 200.0; for(int i = 0; i < 2; ++i){ cin >> a; if(a == ‘E‘) return 0; if(a == ‘T‘){ cin >> T; } else if(a == ‘D‘){ cin >> D; } else if(a == ‘H‘){ cin >> H; } } if(H == 200.0){ H = T + 0.5555 * (6.11 * exp(5417.7530 * (1 / 273.16 - 1 / (D + 273.16))) - 10.0); } else if(T == 200.0){ T = H - 0.5555 * (6.11 * exp(5417.7530 * (1 / 273.16 - 1 / (D + 273.16))) - 10.0); } else if(D == 200.0){ D = 1 / (1 / 273.16 - log(((H - T) / 0.5555 + 10.0) / 6.11) / 5417.7530) - 273.16; } printf("T %.1f D %.1f H %.1f\n", T, D, H); } return 0;}
POJ - 3299 Humidex
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