设答案为r,cnt为x[i] >=0的个数
那么r = 1/n * (Σx[i](x[i] >= 0) + ∑(r - x[i])(x[i] < 0))
然后把r移项到一起解方程, 得到r = ∑|x[i]| / cnt,同除gcd。记得特判下x[i]均为负数的情况即可。
1 #include <cstdio> 2 #include <algorithm> 3 using namespace std; 4 int T,cas,n,tot,gcd,cnt; 5 int x[110]; 6 int main() 7 { 8 for (scanf("%d",&T);T != 0;T--) 9 { 10 cas++; 11 tot = cnt = 0; 12 scanf("%d",&n); 13 for (int i = 1;i <= n;i++) 14 { 15 scanf("%d",&x[i]); 16 if (x[i] >= 0) 17 { 18 cnt++; 19 tot += x[i]; 20 }else 21 tot -= x[i]; 22 } 23 if (cnt == 0) 24 { 25 printf("Case %d: inf\n",cas); 26 continue; 27 } 28 gcd = __gcd(tot,cnt); 29 printf("Case %d: %d/%d\n",cas,tot / gcd,cnt / gcd); 30 31 } 32 return 0; 33 }
原文地址:https://www.cnblogs.com/iat14/p/11410053.html
时间: 2024-11-06 18:24:01