Danganronpa
Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 304 Accepted Submission (s): 217
Problem Descriptionchisa Yukizome works as a teacher in the school. She prepares many gifts, which consist ofNKinds witha[i] Quantities of each kind, for she students and wants to the A class meeting. Because of the busy work, she gives his gifts to the monitor, Chiaki Nanami. Due to the strange design of the school, the students ' desks is in a row. Chiaki Nanami wants to arrange gifts like this:
1. Each table would be prepared to a mysterious gift and an ordinary gift.
2. In order to reflect the Chisa yukizome ' s generosity, the kinds of the ordinary gift on the adjacent table must be Diffe Rent.
3. There is no limits for the mysterious gift.
4. The gift must be placed continuously.
She wants to know what many students can get gifts in accordance with her idea at most (suppose the number of students is Infinite). As the most important people of hers, you're easy-to-solve it, aren ' t you?
Inputthe first line of input contains an integerT(t≤ten) indicating the number of test cases.
Each case contains one integerN. The next line containsN(1≤n≤) Numbersa1,a2,.. . ,an , (1≤ai≤100000).
Outputfor each test case, output one line containing ' case #x: Y ' (without quotes), where x is the ' Test Case number (star Ting from 1) and Y are the answer of Chiaki Nanami ' s question.
Sample Input123 2
Sample outputcase #1:2
Authoruestc
SOURCE2016 Chinese College Student Program design contest-Network Tryouts
When the game is water, it is less than the sum = 1 of the situation WA.
Code:
#include <stdio.h>
#include <string.h>
#define N 110000
int main (void)
{
int sum, I, J, T, N;
int a[n];
j = 0;
scanf ("%d", &t);
while (t--)
{
j + +;
sum = 0;
scanf ("%d", &n);
for (i = 0; i < n; i++)
{
scanf ("%d", &a[i]);
Sum + = A[i];
}
if (sum = = 1)
printf ("Case #%d:1\n", j);
Else
printf ("Case #%d:%d\n", J, SUM/2);
}
}
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Danganronpa water problem.
