链接：http://acm.hdu.edu.cn/search.php?field=problem&key=2012%20Asia%20Chengdu%20Regional%20Contest%20&source=1&searchmode=source

A题(HDU 4464)：无意义的题。。就贴个代码吧。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int maxn = 10010;

int n, cas;
char ch[maxn];

int main() {
	cas = 0;
	while (~scanf("%d", &n)) {
		int ans = 0;
		for (int i = 0; i < n; i++) {
			scanf("%s", ch);
			int len = strlen(ch);
			int sum = 0;
			for (int i = 0; i < len; i++) {
				sum += (int)ch[i];
			}
			if (ans < sum)
				ans = sum;
		}
		printf("Case %d: %d\n", ++cas, ans);
	}
	return 0;
}

B题(HDU 4465)：这题是道求期望的题，还算比较简单。公式很容易得到：

但是因为组合数比较大，不能直接算。为了保证精度，可以先求log，然后再用乘方，算回来。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;

int n, cas = 0;
double p;

int main() {
	while (~scanf("%d %lf", &n, &p)) {
		printf("Case %d: ", ++cas);
		double q = 1 - p;
		if (fabs(p) < 1e-10 || fabs(q)<1e-10) {
			printf("%.6lf\n", 1.0 * n);
			continue;
		}
		double tmp = (n + 1) * log(p);
		double tem = (n + 1) * log(q);
		double ans = 0, t = 0, tt = 0, ttt = 0;
		for (int i = n + 1; i <= n + n; i++) {
			tt = t + tmp + (i-1-n) * log(q) + log(1.0+n+n-i);
			ttt = t + tem + (i-1-n) * log(p) + log(1.0+n+n-i);
			ans += exp(tt) + exp(ttt);
			t += log(1.0*i) - log(1.0*i-n);
		}
		printf("%.6lf\n", ans);
	}
	return 0;
}