安徽的一道大水题,真正的纯模拟,不需要任何技巧,和去年的NOIP一样……
#include <vector>
#include <list>
#include <limits.h>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <memory.h>
using namespace std;
int n,s[1100][40],t[40][1100];
void init()
{
memset(s,0,sizeof(s));
memset(t,0,sizeof(t));
scanf("%d",&n);
for (int i = 1; i <= n; ++ i)
for (int j = 1; j <= 3; ++ j)
{
scanf("%d",&s[i][j]);
++ t[j][ s[i][j] ];
}
}
void work()
{
int ans;
for (int i = 1; i <= n; ++ i)
{
ans = 0;
for (int j = 1; j <= 3; ++ j)
if (t[j][ s[i][j] ] == 1)
ans += s[i][j];
printf("%d\n",ans);
}
}
int main()
{
init();
work();
return 0;
}