本文主要是介绍2013 南京邀请赛 A,C,H, K，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
A题：求期望，先统计出每次得分期望，然后如果获得再一次机会，下次期望为n / m * 期望，n为总个数，m为在一次的个数，然后利用等比数列前
n项和的公式去算即可，要注意特判一下可能出现inf的情况，即期望不为0，且n / m = 1。
C题：a，b范围内，把每个数字拆成32位，计算0-a每位多少个1，0-b每位多少个1，相减后在去计算总进位次数即可
H题：签到题，找出重复数字。
K题：数论，求是否存在ID % X1 属于[Y1,Z1] && ID % X2 属于[Y2, Z2], 设ID % X1 = u, ID % X2 = v, ID / X1 = a, ID / X2 = b; 那么Id - x1 * a = u, id - x2 * b = v;
x2 * b - x1 * a = u - v; 这样就变成了a * x + b *y = c判断有无整数解，欧几里德定理，只要c是gcd(a,b)的整数倍即可。
那么t = gcd(x1, x2)， u - v的范围可以求出来为[y1 - z2, z1 - y2],然后判断一下是不是整数倍即可
代码:
A:
#include <cstdio>
#include <cstring>
#include <cmath>
const double esp = 1e-9;
const int N = 205;int n, m, tmp, vis[N];
double num, sum, mm;int main() {while (~scanf("%d", &n)) {sum = 0; mm = 0;memset(vis, 0, sizeof(vis));for (int i = 0; i < n; i++) {scanf("%lf", &num);sum += num;}sum /= n;scanf("%d", &m);while (m--) {scanf("%d", &tmp);if (!vis[tmp]) mm++;vis[tmp] = 1;}mm /= n;if (fabs(sum) < esp)printf("0.00\n");else if (fabs(sum) >= esp && fabs(mm - 1) < esp)printf("inf\n");else printf("%.2lf\n", sum / (1 - mm));}return 0;
}C:
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int N = 35;
long long a, b;
long long d1[N], d2[N], mi[N];void init() {mi[0] = 1;for (int i = 1; i <= 32; i++)mi[i] = mi[i - 1] * 2;
}void build(long long *d, long long num) {for (int i = 0; i < 32; i++) {long long c = num / mi[i + 1];long long yu = num % mi[i + 1] - mi[i];if (yu < 0) yu = 0;d[i] = c * mi[i] + yu;}
}int main() {init();while (cin >> a >> b) {memset(d1, 0, sizeof(d1));memset(d2, 0, sizeof(d2));build(d1, a);build(d2, b + 1);for (int i = 0; i < 32; i++)d2[i] -= d1[i];long long s = 0, ans = 0;for (int i = 0; i < 32; i++) {s = (d2[i] + s) / 2;ans += s;}while (s) {s /= 2;ans += s;}cout << ans << endl;}return 0;
}H:
#include <stdio.h>
#include <string.h>const int N = 1005;
int n, num, vis[N];int main() {while (~scanf("%d", &n)) {memset(vis, 0, sizeof(vis));for (int i = 0; i <= n; i++) {scanf("%d", &num);vis[num]++;if (vis[num] == 2)printf("%d\n", num);}}return 0;
}K:
#include <stdio.h>
#include <string.h>const int N = 1005;
int n, x[N], y[N], z[N];int gcd(int a, int b) {if (b == 0) return a;return gcd(b, a % b);
}bool judge(int t, int l, int r) {if (l % t == 0 || r % t == 0) return true;if (l < 0 && r >= 0) return true;if (r / t - l / t > 0) return true;return false;
}bool solve() {for (int i = 0; i < n; i++) {for (int j = i + 1; j < n; j++) {int t = gcd(x[i], x[j]);if (judge(t, y[i] - z[j], z[i] - y[j]))return true;}}return false;
}int main() {while (~scanf("%d", &n)) {for (int i = 0; i < n; i++)scanf("%d%d%d", &x[i], &y[i], &z[i]);if (solve())printf("Cannot Take off\n");elseprintf("Can Take off\n");}return 0;
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