Codeforces Round #427 (Div. 2) A. Key races B. The number on the board C. Star sky

A.传送门

B.传送门

A.题意：双方需要输入一个s长度大小的字。然后每个人有一个ping就是延迟，在输入之前和输入完之后提交会延迟两次，问谁快

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e4+100;
const double EPS=1e-7;
const int MOD=10000007;
#define mod(x) ((x)%MOD);
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
inline int Scan()
{
int Res=0,ch,Flag=0;
if((ch=getchar())=='-')Flag=1;
else if(ch>='0' && ch<='9')Res=ch-'0';
while((ch=getchar())>='0'&&ch<='9')Res=Res*10+ch-'0';
return Flag ? -Res : Res;
}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.out" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

LL s,v1,v2,t1,t2;
int check()
{
LL x=t1*2;
LL y=t2*2;
x+=v1*s;
y+=v2*s;
if(x>y)
return 1;
else if(x<y)
return 2;
else return 3;
}
int main()
{
cin>>s>>v1>>v2>>t1>>t2;
if(check()==2)
puts("First");
if(check()==1)
puts("Second");
if(check()==3)
puts("Friendship");
}

B.题意：原数每位数字之和大于k，然后现在把原数变成n，位数不变。（就是改动了几个数字）求原数和n最小的差异

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e3+10;
const int maxx=1e7+100;
const double EPS=1e-7;
const int MOD=10000007;
#define mod(x) ((x)%MOD);
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
inline int Scan()
{
int Res=0,ch,Flag=0;
if((ch=getchar())=='-')Flag=1;
else if(ch>='0' && ch<='9')Res=ch-'0';
while((ch=getchar())>='0'&&ch<='9')Res=Res*10+ch-'0';
return Flag ? -Res : Res;
}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.out" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int k;
string s;
int main()
{
cin>>k;
LL t = 0;
cin >> s;
int len=s.length();
for(int i=0;i<len;i++)
t+=s[i]-'0';
sort(s.begin(),s.end());
if(t>=k)
puts("0");
else
{
LL tm = k-t;
LL ans = 0;
for(int i=0;i<len;i++)
{
ans++;
tm-=9-(s[i]-'0');
if(tm<=0) break;
}
cout <<ans<<endl;
}

}

C.看这里