usaco3.4.2 Electric Fence

一 原题

Electric Fence
Don Piele

In this problem, `lattice points' in the plane are points with integer coordinates.

In order to contain his cows, Farmer John constructs a triangular electric fence by stringing a "hot" wire from the origin (0,0) to a lattice point [n,m] (0<=;n<32,000, 0<m<32,000), then to a lattice point on the positive x axis [p,0] (0<p<32,000), and then back to the origin (0,0).

A cow can be placed at each lattice point within the fence without touching the fence (very thin cows). Cows can not be placed on lattice points that the fence touches. How many cows can a given fence hold?

PROGRAM NAME: fence9

INPUT FORMAT

The single input line contains three space-separated integers that denote n, m, and p.

SAMPLE INPUT (file fence9.in)

7 5 10

OUTPUT FORMAT

A single line with a single integer that represents the number of cows the specified fence can hold.

SAMPLE OUTPUT (file fence9.out)

20

二 分析

PICK定理：直角坐标系中的一个整点多边形的面积S，内部（不含边上）整点数a，边上整点数b三者满足S = a - 1 + b / 2

直角坐标系中端点为(0, 0)和(n, m)的线段上有gcd(n,m)+1个整点

三 代码

运行结果：

USER: Qi Shen [maxkibb3]
TASK: fence9
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 4180 KB]
   Test 2: TEST OK [0.000 secs, 4180 KB]
   Test 3: TEST OK [0.000 secs, 4180 KB]
   Test 4: TEST OK [0.000 secs, 4180 KB]
   Test 5: TEST OK [0.000 secs, 4180 KB]
   Test 6: TEST OK [0.000 secs, 4180 KB]
   Test 7: TEST OK [0.000 secs, 4180 KB]
   Test 8: TEST OK [0.000 secs, 4180 KB]
   Test 9: TEST OK [0.000 secs, 4180 KB]
   Test 10: TEST OK [0.000 secs, 4180 KB]
   Test 11: TEST OK [0.000 secs, 4180 KB]
   Test 12: TEST OK [0.000 secs, 4180 KB]

All tests OK.

Your program ('fence9') produced all correct answers! This is your submission #6 for this problem. Congratulations!

AC代码：

/*
ID:maxkibb3
LANG:C++
PROB:fence9
*/

#include<cstdio>

int n, m, p;

void swap(int *a, int *b) {
    int tmp = *a;
    *a = *b;
    *b = tmp;
}

int gcd(int a, int b) {
    if(a < b) swap(&a, &b);
    if(b == 0) return a;
    else return gcd(b, a % b);
}

int main() {
    freopen("fence9.in", "r", stdin);
    freopen("fence9.out", "w", stdout);
    scanf("%d%d%d", &n, &m, &p);

    int S, a, b = 0;
    S = p * m / 2;
    b += gcd(n, m) + gcd(abs(p - n), m) + p;
    a = S + 1 - b / 2;

    printf("%d\n", a);
    return 0;
}