http://acm.hdu.edu.cn/showproblem.php?pid=4974
话说是签到题,我也不懂什么是签到题。
A simple water problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 196 Accepted Submission(s):
124
competition is held between two competitors, and surely Dragon's favorite. After
each competition he will give a score of either 0 or 1 for each competitor and
add it to the total score of that competitor. The total score begins with zero.
Here's an example: four competitors with name James, Victoria, Penghu, and Digo.
First goes a competition between Penghu and Digo, and Dragon enjoys the
competition and draw both 1 score for them. Then there’s a competition between
James and Victoria, but this time Dragon draw 1 for Victoria and 0 for James.
Lastly a competition between James and Digo is held, but this time Dragon really
dislike the competition and give zeroes for each of them. Finally we know the
score for each one: James--0, Victoria--1, Penghu--1, Digo--1. All except James
are the Winner!
However, Dragon's mom comes back home again and close the
TV, driving Dragon to his homework, and find out the paper with scores of all
competitors. Dragon's mom wants to know how many competition Dragon watched, but
it's hard through the paper. Here comes the problem for you, given the scores of
all competitors, at least how many competitions had Dragon watched?
T(<=10), the number of test cases. Following T blocks, each block describe
one test case.
For each test case, the first line contains only one
integers N(<=100000), which means the number of competitors. Then a line
contains N integers
(a1,a2,a3,...,an).ai(<=1000000)
means the score of i-th competitor.
start with "Case #i: ", with i implying the case number. Then for each case just
puts a line with one integer, implying the competition at least should be
watched by dragon.
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[];
int main()
{
int n,t,k=,i,sum,mmax;
scanf("%d",&t);
while(t--)
{
sum=mmax=;
scanf("%d",&n);
for(i=;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
if(mmax<a[i])
mmax=a[i];
}
int ans=max((sum+)/,mmax);
printf("Case #%d: %d\n",k++,ans);
}
return ;
}