Codeforces Round #441 (Div. 2, by Moscow Team Olympiad) ABCD题解

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A.

题意：

兔子和猫头鹰有一条路距离为a，兔子和屹耳驴有一条路距离为b，猫头鹰和屹耳驴有一条路距离为c，

现在小熊在兔子家吃了饭，他要一共要吃n顿饭，问怎么走才能走最少的路程。每次离开这个主人的家

之后，这个主人又做了新饭。

pair 搞搞。

/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#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 s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#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 fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
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=4e3+10;
const int maxx=2e6+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
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));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

int a,b,c;
int n;
ii s(int x)
{
if(x==1)
{
if(a<b) return mp(a,2);
else return mp(b,3);
}
if(x==2)
{
if(a<c) return mp(a,1);
else return mp(c,3);
}
if(x==3)
{
if(b<c) return mp(b,1);
else return mp(c,2);
}
}
void solve()
{
s_1(n);
s_3(a,b,c);
int pre=1,ans=0;
n--;
W(n--)
{
ii z=s(pre);
ans+=z.first;
pre=z.second;
}
print(ans);
}

int main()
{
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}


B.
n个数，让你选k个数，他们同余m。

暴力。

头文件一样就不多ctrl c了。

int n,k,m;

int a[maxn];
int b[maxn];
void solve()
{
s_3(n,k,m);
FOR(1,n,i)
s_1(a[i]);
FOR(1,n,i)
b[a[i]%m]++;
int cal=-1;
FOr(0,m,i)
{
if(b[i]>=k)
{
cal=i;
break;
}
}
if(cal==-1)
puts("No");
else
{
puts("Yes");
int cnt=0;
FOR(1,n,i)
{
if(a[i]%m==cal)
{
cnt++;
cout<<a[i]<<" ";
}
if(cnt==k)
break;
}
}
}

int main()
{
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}

C.
给你一个n，问1-n内有多少x，x的每一位加起来+x=n。

看样例就知道了，

每一位最多9，有多少位呢，8位吧，最多72，所以还是暴力。

我人傻写dfs，没交上去，不知道会不会T，应该优秀的dfs可以过

int n;
int a[maxn];
int ans=0;
int calc(int x)
{
int sum=x;
int t=x;
W(t)
{
sum+=t%10;
t/=10;
}
return sum;
}
void solve()
{
s_1(n);
int x=n-100;
FOR(max(0,x),n,i)
{
if(calc(i)==n)
{
a[++ans]=i;
}
}
print(ans);
FOR(1,ans,i)
{
print(a[i]);
}
}

int main()
{
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}

D.
找最高位置的X，然后看前面的O在哪个位置。

int ans=1;
int a[maxn];
int x[maxn];
int n;
void solve()
{
s_1(n);
FOR(1,n,i)
s_1(a[i]);
int r=n;
printf("1");
FOR(1,n-1,i)
{
x[a[i]]=1;
if(x[r]==1)
{
W(x[r]==1) r--,ans--;
ans++;
}
if(a[i]<r) ans++;
printf(" %d",ans);
}
printf(" 1\n");
}

int main()
{
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}