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

传送门

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_backtypedef 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();    }}