A. Contest
time limit per test
memory limit per test
input
output
a points and Vasya solved the problem that costs b points. Besides, Misha submitted the problem c minutes after the contest started and Vasya submitted the problem d minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs p points tminutes after the contest started, you get
points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input
a, b, c, d (250 ≤ a, b ≤ 3500, 0 ≤ c, d ≤ 180).
a and b are divisible by 250
Output
Output on a single line:
Misha" (without the quotes), if Misha got more points than Vasya.
Vasya" (without the quotes), if Vasya got more points than Misha.
Tie" (without the quotes), if both of them got the same number of points.
Sample test(s)
input
500 1000 20 30
output
Vasya
input
1000 1000 1 1
output
Tie
input
1500 1000 176 177
output
Misha
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
int main()
{
__int64 a,b,c,d;
while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&d)!=EOF)
{
__int64 x,y;
x = max((3*a/10),a-(a/250)*c);
y = max((3*b/10),b-(b/250)*d);
if(x>y)
{
printf("Misha\n");
}
else if(x<y)
{
printf("Vasya\n");
}
else
{
printf("Tie\n");
}
}
return 0;
}
B. Misha and Changing Handles
time limit per test
memory limit per test
input
output
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
q (1 ≤ q ≤ 1000), the number of handle change requests.
q
old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
old, and handlenew
Output
n
n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old andnew, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Sample test(s)
input
5Misha ILoveCodeforces Vasya Petrov Petrov VasyaPetrov123 ILoveCodeforces MikeMirzayanov Petya Ivanov
output
3Petya Ivanov Misha MikeMirzayanov Vasya VasyaPetrov123
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<map>
using namespace std;
char a[10100][21];
char b[10010][21];
struct node
{
char x[21];
char y[21];
}p[10010];
int cmp1(const void *a,const void *b)
{
struct node *aa,*bb;
aa = (struct node*)a;
bb = (struct node*)b;
return strcmp(aa->y,bb->y);
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
map<string,int>q;
char str[21];
for(int i=1;i<=n;i++)
{
scanf("%s%s",b[i],a[i]);
q[b[i]] = i;
}
for(int i=1;i<=n;i++)
{
if(q[b[i]]!=0)
{
while(q[a[q[b[i]]]]!=0)
{
strcpy(str,a[q[b[i]]]);
q[b[i]] = q[a[q[b[i]]]];
q[str] = 0;
}
}
}
int k = 0;
for(int i=1;i<=n;i++)
{
if(q[b[i]]!=0)
{
strcpy(p[k].x,b[i]);
strcpy(p[k].y,a[q[b[i]]]);
k++;
}
}
printf("%d\n",k);
qsort(p,k,sizeof(p[0]),cmp1);
for(int i=0;i<k;i++)
{
printf("%s %s\n",p[i].x,p[i].y);
}
}
return 0;
}
