ACM-PKU
PKU 题号1004，题目Financial Management
求12个浮点数的平均数，比较水的题目。注意打印的时候，“%.2f”C语言自动完成四舍五入了。
#include<stdio.h> #define ONLINE void online() { #ifdef ONLINE #else freopen("Financial.in", "r", stdin); freopen("Financial.out", "w", stdout); #endif } int main() { online(); float sum = 0.000; float month = 0.000; for (int i = 0; i < 12; i ++) { scanf("%f", &month); sum += month; } sum = sum / 12; printf("$%.2f", sum); return 0; }
题号：1005，题目I Think I Need a Houseboat
以原点为中心的半圆进行腐蚀，每年的腐蚀面积为50平米，问点P能够支撑多少年不被腐蚀到。可以计算以原点O到P为半径的半圆的面积，然后除以50每年，即得到年数，取整+1，也是水题。
#include <stdio.h> #define ONLINE void online() { #ifdef ONLINE #else freopen("houseboat.in","r", stdin); freopen("houseboat.out", "w", stdout); #endif } int n; float x, y; const float PI = 3.1415926; int main() { online(); scanf("%d", &n); for (int i=1; i <= n; i ++) { scanf("%f%f", &x, &y); float year = (x * x + y * y) * PI / 100; printf("Property %d: This property will begin eroding in year %d./n", i, (int)year + 1); } printf("END OF OUTPUT./n"); return 0; }
题号：1006，题目Biorhythms
这道题要好好说说，具体可以看解题报告，http://zi-jin.com/archives/52，写的很好，但是有个地方理解错了。具体可以看，中国剩余定理http://mba.shengwushibie.com/itbook/BookChapter.asp?id=17863
#include <stdio.h> //#define ONLINE void online() { #ifdef ONLINE #else freopen("bio.in", "r", stdin); freopen("bio.out", "w", stdout); #endif } const int m[3] = {23, 28, 33}; int _gcd, t; void gcd(int a, int b, int &x, int &y) { if (!b) { x = 1; y = 0; _gcd = a; return; } gcd(b, a % b, x, y); t = x; x = y; y = t - a / b * y; } int solve(int r[], int k, int &M) { //m[i]同文中的m_i //k同文中的k，本题是定值3,但其他题目中可能是变量 int i, M_i, x, y, a = 0; for (i = 0, M = 1; i < k; M *= m[i++]); for (i = 0; i < k; ++i) { M_i = M / m[i]; //求逆的过程 gcd(m[i], M_i, x, y); a = (a + y * M_i * r[i]) % M; } return (a + M) % M; //保证a非负 } int main() { online(); int r[3], d=0, c = 1, ans, M; while (scanf("%d%d%d%d", &r[0], &r[1], &r[2], &d) != EOF && r[0] != -1) { r[0] %= m[0]; r[1] %= m[1]; r[2] %= m[2]; ans = solve(r, 3, M); while (ans <= d) ans += M; printf("Case %d: the next triple peak occurs in %d days./n", c, ans - d); ++c; } }