Google Code Jam 2014 A,B,D

A题：水题，统计两列上数字要是只有一个重复输出该数字，如果有多个，就是bad，如果没有就是cheat

B题：推了个数学公式，y(n) = c / 2 + c / (f + 2) + c /(2 * f + 2) + ... + c / ((n - 1) * f + 2) + x / (n * f + 2);

然后y(n) - y(n - 1)化简后得到分母>0,分子为c * n * f + 2 * c - x * f，可以看出函数是先单调递减在单调递增，要求最小值n = (x * f - 2 * c) / (c * f) （上取整和下取整的最小值），然后for一遍计算出答案即可

D题：欺诈游戏就用田忌赛马的方法，真实游戏就用常规的方法即可

C题：扫雷，，还没想到什么好方法，感觉像构造，如果暴力只能过小数据

代码：

A：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

const int N = 5;
int t, s1, s2, i, j;
int g1[N][N], g2[N][N], vis[20];

void solve() {
	int sum = 0;
	int v;
	for (int i = 1; i <= 16; i++) {
		if (vis[i]) sum ++;
		if (vis[i] == 2) v = i;
	}
	if (sum < 7) printf("Bad magician!\n");
	else if (sum > 7) printf("Volunteer cheated!\n");
	else printf("%d\n", v);
}

int main() {
	int cas = 0;
	scanf("%d", &t);
	while (t--) {	
		memset(vis, 0, sizeof(vis));
		scanf("%d", &s1);
		for (i = 0; i < 4; i ++)
			for (j = 0; j < 4; j++)
				scanf("%d", &g1[i][j]);
		scanf("%d", &s2);
		for (i = 0; i < 4; i ++)
			for (j = 0; j < 4; j++)
				scanf("%d", &g2[i][j]);
		for (i = 0; i < 4; i++)
			vis[g1[s1 - 1][i]]++;
		for (i = 0; i < 4; i++)
			vis[g2[s2 - 1][i]]++;
		printf("Case #%d: ", ++cas);
		solve();
	}
	return 0;
}

#include <stdio.h>
#include <string.h>
#include <math.h>
#define min(a, b) ((a)<(b)?(a):(b))
int t;
double c, f, x;

int main() {
	int cas = 0;
	scanf("%d", &t);
	while (t--) {
		scanf("%lf%lf%lf", &c, &f, &x);
		int n = (int)floor((f * x - 2 * c) / (c * f));
		double sum1 = 0;
		if (n < 0) n = 0;
		for (int i = 0; i < n; i++) {
			sum1 += c / (f * i + 2);
		}
		double sum2 = sum1 + c / (f * n + 2);
		sum1 += x / (f * n + 2);
		sum2 += x / (f * (n + 1) + 2);
		double sum = min(sum1, sum2);
		printf("Case #%d: %.7lf\n", ++cas, sum);
	}
	return 0;
}

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int N = 1005;
int t, n, i, j, ans1, ans2;
double a1[N], a2[N];

int main() {
	int cas = 0;
	scanf("%d", &t);
	while (t--) {
		ans1 = ans2 = 0;
		scanf("%d", &n);
		for (i = 0; i < n; i++)
			scanf("%lf", &a1[i]);
		for (i = 0; i < n; i++)
			scanf("%lf", &a2[i]);
		sort(a1, a1 + n);
		sort(a2, a2 + n);
		i = 0; j = 0;
		while (i != n && j != n) {
			if (a1[i] > a2[j])
				j++;
			else {
				i++; j++;
			}
		}
		ans2 = j - i;
		int s = 0, e = n - 1;
		j = n - 1;
		while (s <= e) {
			if (a1[e] < a2[j]) {
				s++; j--;
			}
			else {
				e--; j--;
				ans1++;
			}
		}
		printf("Case #%d: %d %d\n", ++cas, ans1, ans2);
	}
	return 0;
}

Google Code Jam 2014 A,B,D,布布扣,bubuko.com

Google Code Jam 2014 A,B,D

原文：http://blog.csdn.net/accelerator_/article/details/23598635