标签:acm区域赛 hdu algorithm 成都 icpc
Browsing History
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3065 Accepted Submission(s): 1692
Problem Description
One day when you are going to clear all your browsing history, you come up with an idea: You want to figure out what your most valued site is. Every site is given a value which equals to the sum of ASCII values of all characters in the URL. For example aa.cc
has value of 438 because 438 = 97 + 97 + 46 + 99 + 99. You just need to print the largest value amongst all values of sites.
Things are simplified because you found that all entries in your browsing history are of the following format: [domain], where [domain] consists of lower-case Latin letters and “.” only. See the sample input for more details.
Input
There are several test cases.
For each test case, the first line contains an integer n (1 ≤ n ≤ 100), the number of entries in your browsing history.
Then follows n lines, each consisting of one URL whose length will not exceed 100.
Input is terminated by EOF.
Output
For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is a number indicating the desired answer.
Sample Input
1
aa.cc
2
www.google.com
www.wikipedia.org
Sample Output
Source
水题不解释....
AC代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
int n, k = 1;
char a[105];
while(~scanf("%d", &n))
{
int ans=0;
n--;
scanf("%s", a);
for(int i=0; i<strlen(a); i++)
{
ans+=a[i];
}
while(n--)
{
scanf("%s", a);
int tmp = 0;
for(int i = 0; i<strlen(a); i++)
{
tmp += a[i];
}
if(tmp>ans) ans = tmp;
}
printf("Case %d: %d\n", k++, ans);
}
return 0;
}
HDU-4464-Browsing History (2012 ACM/ICPC成都现场赛!)
标签:acm区域赛 hdu algorithm 成都 icpc
原文地址:http://blog.csdn.net/u014355480/article/details/41521745