题目简单，主要是对名次的处理。下面的代码巧妙的对名次 那一块做了处理。如果 三个同学的 最长距离分别是10 20 30，然后 分别对应的是x[1],x[2],x[3];分别比较，当x[i]>x[j] 时，y[i]++，当所有的成绩都比较完之后，再对y[i]+1即得到 其排名。
Jump and Jump...
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1334 Accepted Submission(s): 677
Problem Description
There are n kids and they want to know who can jump the farthest. For each kid, he can jump three times and the distance he jumps is maximum distance amount all the three jump. For example, if the distance of each jump is (10, 30, 20), then the farthest distance he can jump is 30. Given the distance for each jump of the kids, you should find the rank of each kid.
Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤100), indicating the number of test cases. For each test case: The first line contains an integer n (2≤n≤3), indicating the number of kids. For the next n lines, each line contains three integers ai,bi and ci (1≤ai,bi,ci,≤300), indicating the distance for each jump of the i-th kid. It‘s guaranteed that the final rank of each kid won‘t be the same (ie. the farthest distance each kid can jump won‘t be the same).
Output
For each test case, you should output a single line contain n integers, separated by one space. The i-th integer indicating the rank of i-th kid.
Sample Input
2 3 10 10 10 10 20 30 10 10 20 2 3 4 1 1 2 1
Sample Output
3 1 2 1 2HintFor the first case, the farthest distance each kid can jump is 10, 30 and 20. So the rank is 3, 1, 2.
Source
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
int a,b,c,t,n,x[4],i,j,y[4];
scanf("%d",&t);
while (t--)
{
scanf("%d",&n);
for (i=1;i<=n;i++)
{
scanf("%d%d%d",&a,&b,&c);
x[i]=max(max(a,b),c);
}
memset(y,0,sizeof(y));
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
if (x[i]<x[j]) y[i]++; /*对名次进行处理*/
}
for (i=1;i<n;i++) printf("%d ",y[i]+1);
printf("%d\n",y[n]+1);
}
return 0;
}