CodeForces#285 div.2 题解

A题：

题意：给出两个人的提交时间和题目的初始分数，比较两人的成绩的高低

思路：根据公式模拟即可max(3*p/10,p-p/250*t)

下面是A题代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);

int main()
{
ios::sync_with_stdio( false );

int a, b, c, d;
cin >> a >> b >> c >> d;
int ans_1, ans_2;
ans_1 = max ( 3 * a / 10, a - a / 250 * c );
ans_2 = max ( 3 * b / 10, b - b / 250 * d );
if ( ans_1 > ans_2 ){
cout << "Misha" << endl;
}
else if ( ans_1 < ans_2 ){
cout << "Vasya" << endl;
}
else if ( ans_1 == ans_2 ){
cout << "Tie" << endl;
}

return 0;
}

B题：

题意：给出一个数字n，下面有n组数据，每组数据表示一个人的旧名称和新名称，按新名称升序排序，输出初始名称(最早)和当前名称(最新)。

思路：暴力查询、更新即可

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);

struct name{
string a, b;
name(){}
name(string a, string b){
this->a = a;
this->b = b;
}
friend bool operator < ( const name &a, const name &b ){
return a.b < b.b;
}
};

int main()
{
ios::sync_with_stdio( false );
vector<name> num;
int n;
cin >> n;
string a, b;
cin >> a >> b;
num.push_back ( name(a,b) );
for ( int i = 1; i < n; i ++ ){
cin >> a >> b;
int j;
for ( j = 0; j < num.size(); j ++ ){
if ( a == num[j].b ){
num[j].b = b;
break;
}
if ( b == num[j].a ){
num[j].a = a;
break;
}
}
if ( j == num.size() ){
num.push_back ( name( a, b ) );
}
}
sort ( num.begin(), num.end() );
cout << num.size() << endl;
for ( int i = 0; i < num.size(); i ++ ){
cout << num[i].a << ' ' << num[i].b << endl;
}
return 0;
}

C题：

(未完待续)