POJ 1005-I Think I Need a Houseboat
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每年的土地流失面积一定,但是流失的土地的半径增长却是逐渐变缓的。
即第n年的流失的土地半径满足关系 :rn ^ 2 = 100n / pai
可以变换一下形式,即 n = pai * rn ^ 2 / 100
而某点到圆心的距离的平方为 d ^ 2 = x ^ 2 + y ^ 2
显然,n年是满足系数为1的正比例函数的。
所以当给定了一个点的坐标时,根据该坐标所求得的距离圆心的长度二算出的n,就是土地流失的时间。
而由于年份应该取正整数,所以需要进行一下向上取整处理。
即 n = [d ^ 2 * pai / 100] + 1
#include <stdio.h>#include <stdlib.h>#include <math.h>#define PAI 3.1415926int main(){ int N, i, r, n; float x, y; float a; scanf("%d", &N); for(i = 0; i < N; i++) { scanf("%f%f", &x, &y); n = i + 1; //a = sqrt(((x * x + y * y) * PAI / n) / 10); a = (x * x + y * y) * PAI / 100; printf("Property %d: This property will begin eroding in year %d.\n", n, ((int)a + 1)); } printf("END OF OUTPUT."); return 0;}
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