J. Factorial Products
思路:根據對數性質:log(a*b) = log(a) + log(b)，使得階乘相乘轉變為前綴和累加，處理了資料過大無法存盤的問題，
#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> using namespace std; const double eps = 1e-7; const int N = 2600; double pre[N]; void init(){ for(int i = 2; i <= 2560; ++i) pre[i] = pre[i - 1] + (log(i*1.0)); } void solve(){ init(); int T; scanf("%d", &T); for(int t = 1; t <= T; ++t){ int a, b, c, x; double aa = 0, bb = 0, cc = 0; scanf("%d%d%d", &a, &b, &c); for(int i = 1; i <= a; ++i){ scanf("%d", &x); aa += pre[x]; } for(int i = 1; i <= b; ++i){ scanf("%d", &x); bb += pre[x]; } for(int i = 1; i <= c; ++i){ scanf("%d", &x); cc += pre[x]; } printf("Case #%d: ", t); if(fabs(aa - bb) <= eps && fabs(aa - cc) <= eps && fabs(cc - bb) <= eps){ printf("TIE\n"); } else if(fabs(aa - bb) <= eps){ if(cc - aa > 0) printf("C\n"); else printf("TIE\n"); }else if(fabs(aa - cc) <= eps){ if(bb - aa > 0) printf("B\n"); else printf("TIE\n"); }else if(fabs(bb - cc) <= eps){ if(aa - bb > 0) printf("A\n"); else printf("TIE\n"); }else{ if(aa - bb > 0 && aa - cc > 0) printf("A\n"); else if(bb - aa > 0 && bb - cc > 0) printf("B\n"); else if(cc - aa > 0 && cc - bb > 0) printf("C\n"); } } } int main(){ solve(); return 0; }
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