http://acm.hdu.edu.cn/showproblem.php?pid=4972
++和+1还是有区别的，不可大意。
A simple dynamic programming problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 307 Accepted Submission(s):
117
Heat.
Here's an introduction of basketball
game:http://en.wikipedia.org/wiki/Basketball. However the game in Dragon's
version is much easier:
"There's two teams fight for the winner. The only
way to gain scores is to throw the basketball into the basket. Each time after
throwing into the basket, the score gained by the team is 1, 2 or 3. However due
to the uncertain factors in the game, it’s hard to predict which team will get
the next goal".
Dragon is a crazy fan of Miami Heat so that after each
throw, he will write down the difference between two team's score regardless of
which team keeping ahead. For example, if Heat's score is 15 and the opposite
team's score is 20, Dragon will write down 5. On the contrary, if Heat has 20
points and the opposite team has 15 points, Dragon will still write down
5.
Several days after the game, Dragon finds out the paper with his
record, but he forgets the result of the game. It's also fun to look though the
differences without knowing who lead the game, for there are so many uncertain!
Dragon loves uncertain, and he wants to know how many results could the
game has gone?
the number of test cases. Following T blocks, each block describe one test
case.
For each test case, the first line contains only one integer
N(N<=100000), which means the number of records on the paper. Then there
comes a line with N integers (a1, a2, a3, ... ,
an). ai means the number of i-th record.
start with "Case #i: ", with i implying the case number. Then for each case just
puts an integer, implying the number of result could the game has
gone.
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int abs(int x)
{
if(x>)
return x;
else
return -x;
}
int main()
{
int i,t,n,a[],ans,k=;
scanf("%d",&t);
while(t--)
{
memset(a,,sizeof(a));
ans=;
scanf("%d",&n);
for(i=;i<n;i++)
scanf("%d",&a[i]);
int cnt=;
int flag=;
for(i=;i<n&&(i+)<n;i++)
{
if(((a[i+]==a[i])&&a[i]!=)||abs(a[i+]-a[i])>)
{
flag=;
break;
}
if((a[i]==&&a[i+]==)||(a[i]==&&a[i+]==))
cnt++;
} if(flag==)
{
printf("Case #%d: %d\n",k++,ans);
continue;
}
if(a[n-]==)
ans=cnt+;
else
ans=*cnt+;
printf("Case #%d: %d\n",k++,ans);
}
return ;
}