Codeforces Round #451 (Div. 2) 划水报告

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A.Rounding
题意：给出一个数x,求出和x差值绝对值最小的,%10==0的数
贪心向上取整向下取整即可

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int A[200000+10];int main(){    //freopen("a.in","r",stdin);    //freopen("a.out","w",stdout);    int n;    scanf("%d",&n);    if(n%10==0)        printf("%d\n",n);    else if(n%10<=5)        printf("%d\n",(n/10)*10);    else printf("%d\n",(n/10+1)*10);return 0;}

B. Proper Nutrition
给出a,b,n
求a∗x+b∗y==n的非负整数解
裸的扩展欧几里得

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int A[200000+10];typedef long long ll;inline void exgcd(ll a,ll b,ll &d,ll &x,ll &y){    if(!b){        d=a;        x=1;        y=0;        return ;    }    else {        exgcd(b,a%b,d,y,x);        y-=(a/b)*x;    }}inline ll ab(ll x){    return x<0?-x:x;}int main(){    //freopen("a.in","r",stdin);    //freopen("a.out","w",stdout);    ll n,a,b;    scanf("%lld %lld %lld",&n,&a,&b);    ll d,x,y;    exgcd(a,b,d,x,y);    //printf("%d\n",n%d);    if(n%d!=0){        puts("NO");        return 0;    }    ll aa=a/d,bb=b/d;    x*=(n/d),y*=(n/d);    ll p=(x%bb+bb)%bb;    ll tl=ab(x-p)/bb;    if(x>p)        y+=tl*aa;    else y-=tl*aa;    if(y>=0)        printf("YES\n%lld %lld\n",p,y);    else puts("NO");return 0;}

C. Phone Numbers
给出几段人名以及对应的几个字符串
将人名对应的字符串去重（若a为b的后缀a也算重复）输出
set判重，暴力判后缀

By Ostmbh, contest: Codeforces Round #451 (Div. 2), problem: (C) Phone Numbers, Accepted, # #include <iostream>#include <cstdio>#include <cstring>#include <string>#include <map>using namespace std;map<string,int>m;struct T{    string nme;    string B[310];    int vis[310];    int cnt;    int tt;    T(){        cnt=0;        memset(vis,0,sizeof(vis));    }}A[30];int tot=0;inline bool judge(int x,int y,int z){    int ly=A[x].B[y].length(),lz=A[x].B[z].length();    if(lz>ly)        return false;    int i=1;    while(i<=lz&&A[x].B[y][ly-i]==A[x].B[z][lz-i])        i++;    if(i==lz+1)        return true;    return false;}int main(){    //freopen("a.in","r",stdin);    //freopen("a.out","w",stdout);    int n;    scanf("%d",&n);    int x;    string s;    for(int i=1;i<=n;i++){        cin>>s;        if(!m[s])            m[s]=++tot;        int u=m[s];        A[u].nme=s;        scanf("%d",&x);        for(int j=1;j<=x;j++)            cin>>A[u].B[++A[u].cnt];    }    //printf("%d\n",tot);    for(int i=1;i<=tot;i++){        //printf("%d\n",A[i].cnt);        for(int j=1;j<=A[i].cnt;j++)            for(int k=1;k<=A[i].cnt;k++){                if(j==k)                    continue;                if(judge(i,j,k)&&A[i].B[j]!=A[i].B[k])                    A[i].vis[k]=1;                else if(A[i].B[j]==A[i].B[k]){                    if(!A[i].vis[j]&&!A[i].vis[k])                        A[i].vis[k]=1;                }            }        for(int j=1;j<=A[i].cnt;j++)            if(!A[i].vis[j])                A[i].tt++;    }    printf("%d\n",tot);    for(int i=1;i<=tot;i++){        cout<<A[i].nme<<' '<<A[i].tt<<' ';        for(int j=1;j<=A[i].cnt;j++)            if(!A[i].vis[j])                cout<<A[i].B[j]<<' ';        cout<<endl;    }return 0;}

D. Alarm Clock
给出一串序列
问最少去除多少个数是的数轴上任意[x,x+m]区间内包括的序列中的数少于k个
排序+单调队列维护
在前面留数一定比在后面留优

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;int A[200000+10];queue<int>q;int main(){    //freopen("a.in","r",stdin);    //freopen("a.out","w",stdout);    int n,m,k;    scanf("%d %d %d",&n,&m,&k);    k--;    if(k<=0){        printf("%d\n",n);        return 0;    }    for(int i=1;i<=n;i++)        scanf("%d",&A[i]);    sort(A+1,A+n+1);    int now=2;    q.push(1);    int cnt=1;    int ans=0;    while(now<=n){        int x=q.front();        //printf("%d\n",x);        while(now<=n){            //printf("%d\n",A[now]);            if(A[now]>=A[x]+m){                q.push(now);                cnt++;                now++;                break;            }            else {                if(cnt<k){                    q.push(now);                    cnt++;                }                else ans++;                now++;            }        }        cnt--;        q.pop();    }    printf("%d\n",ans);return 0;}

E. Squares and not squares
给出一个长度为偶数的序列
每次可以把一个数x改为x-1
求最少改多少次可以把这个序列变成一个恰好有一半是完全平方数
另一半不是
贪心先求一边完全平方数数量求出要补什么
然后把补得代价排序贪心取

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <cmath>using namespace std;typedef long long ll;ll A[200000+10];ll dl[200000+10];ll bl[200000+10];ll res[200000+10];inline ll get_up(ll x){    if(x==0)        return 2;    return 1;}inline ll get_sq(ll x){    ll l=0,r=x;    while(l+1<r){        ll mid=(l+r)>>1;        if(mid*mid>=x)            r=mid;        else l=mid;    }    if(r*r<=x)        return r;    return l;}   inline ll get_down(ll x){    if(x==0)        return dl[0];    if(x==1)        return dl[0];    return 1;}int main(){    ////freopen("a.in","r",stdin);    //freopen("a.out","w",stdout);    int n;    scanf("%d",&n);    int nw=0;    memset(bl,127,sizeof(bl));    memset(dl,127,sizeof(dl));    //printf("%lld\n",bl[0]);    for(int i=1;i<=n;i++){        scanf("%lld",&A[i]);        ll td=get_sq(A[i]);        if(td*td==A[i]){            nw++;            bl[i]=min(get_up(td),get_down(td));        }        else {            dl[i]=min(A[i]-td*td,(td+1)*(td+1)-A[i]);//变成完全平方数的代        }    }    int u=n-nw;    if(u==nw){        puts("0");        return 0;    }    long long ans=0;    if(u>nw){        int res=(u-nw)/2;        sort(dl+1,dl+n+1);        for(int i=1;i<=res;i++)            ans+=dl[i];    }    else {        int res=(nw-u)/2;        //printf("%d\n",res);        sort(bl+1,bl+n+1);        for(int i=1;i<=res;i++)            ans+=bl[i];    }    cout<<ans<<endl;return 0;}

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