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  1. #include <stdio.h>
  2.  
  3. int main(void) {
  4.  double hani = powl(10.0L, 19.0L);
  5.  double go = 1.0L;
  6.  double yon = 1.0L;
  7.   int n = 0;
  8.    while (1) {
  9.     n++;
  10.     go *= 5.0L;
  11.     yon *= 4.0L;
  12.      if (go + yon > hani) {
  13.       printf("n = %d で 5^n + 4^n = %.0Lf + %.0Lf = %.0Lf > 10^19\n",
  14.        n, go, yon, go + yon);
  15.    break;
  16.   }
  17.  }
  18. }
Success #stdin #stdout 0.01s 5320KB
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Standard input is empty
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n = 28 で 5^n + 4^n = 0 + 0 = nan > 10^19
https://ideone.com/b8TAk3
language:
C (gcc 8.3)
created:
9 days ago
visibility:
public

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