ZOJ-2545

也比较简单，转化为对数计算就行了

#include<stdio.h>#include<math.h>int main(){int i, a[23], n = 1;double sum = 0, log2 = log(2);for (i = 2; i <= 22; i++){int max = 1 << i;while (sum / log2 < max)sum += log(n++);a[i] = n - 2;}while (scanf("%d", &n), n)printf("%d\n", a[(n - 1960) / 10 + 2]);return 0;}