zoj 1242 Carbon Dating(math)

数学题。题意实在是烦，w是质量，d是t时刻的总质量。求t。d/w=810*2^(-t/T).再用log转换下

四舍五入的话，将其除以100或者1000再加上0.5，强制转换为int就行。

#include<stdio.h>#include<string.h>#include<math.h>int main(){double age;double w,d;int Case=1,year;while(~scanf("%lf%lf",&w,&d),w,d){age=-5730*(log(d/(810*w)))/log(2);if(age<=10000)year=(int)(age/100+0.5)*100;else if(age<=100000)year=(int)(age/1000+0.5)*1000;printf("Sample #%d\nThe approximate age is %d years.\n\n",Case++,year);}return 0;}