contest0 20180804

Posted 2022-05-08 blogofchc1234567890

A

Code

#include<bits/stdc++.h>
using namespace std;
const int maxc=1003,mod=1000000007;
int n,m,c,a[maxc],b[maxc],dp[maxc][maxc];
int Plus(int x,int y)return (x+=y)>=mod?x-mod:x;
void PlusEqual(int &x,int y)if((x+=y)>=mod)x-=mod;
int mul(long long x,int y)return x*y%mod;
int main()
    scanf("%d%d%d",&n,&m,&c);
    for(int i=1;i<=n;i++)
        int x;
        scanf("%d",&x);
        a[x]++;
    
    for(int i=1;i<=m;i++)
        int x;
        scanf("%d",&x);
        b[x]++;
    
    for(int i=c;i>=0;i--)
        a[i]=mul(a[i],b[i]);
        dp[i][0]=Plus(dp[i+1][0],a[i]);
    
    for(int i=c;i>=0;i--)
        for(int j=c;j>=1;j--)
            dp[i][j]=Plus(mul(dp[i+1][j-1],a[i]),dp[i+1][j]);
        
    
    for(int i=1;i<=c;i++)
        printf("%d ",dp[1][i]);
    
    return 0;

B

Code

#include<bits/stdc++.h>
using namespace std;
const int maxn=100003,mod=1000000007;
int n,Q,q[maxn][3],t[maxn],TT[maxn];
int Plus(int x,int y)return (x+=y)>=mod?x-mod:x;
int Minus(int x,int y)return Plus(x,mod-y);
void PlusEqual(int &x,int y)if((x+=y)>=mod)x-=mod;
int mul(long long x,int y)return x*y%mod;
void add(int pos,int k)while(pos<=Q)PlusEqual(t[pos],k),pos+=pos&-pos;
int query(int pos)int ret=0;while(pos)PlusEqual(ret,t[pos]),pos-=pos&-pos;return ret;
void ADD(int pos,int k)while(pos<=n)PlusEqual(TT[pos],k),pos+=pos&-pos;
int QUERY(int pos)int ret=0;while(pos)PlusEqual(ret,TT[pos]),pos-=pos&-pos;return ret;
int main()
    int T;
    scanf("%d",&T);
    while(T--)
        scanf("%d%d",&n,&Q);
        memset(t,0,sizeof(t));
        memset(TT,0,sizeof(TT));
        for(int i=1;i<=Q;i++)
            for(int j=0;j<=2;j++)scanf("%d",&q[i][j]);
            add(i,1),add(i+1,-1);
        
        for(int i=Q;i>=1;i--)
            if(q[i][0]==2)
                int tmp=query(i);
                add(q[i][1],tmp),add(q[i][2]+1,Minus(0,tmp));
            
        
        for(int i=1;i<=Q;i++)
            if(q[i][0]==1)
                int tmp=query(i);
                ADD(q[i][1],tmp),ADD(q[i][2]+1,Minus(0,tmp));
            
        
        for(int i=1;i<=n;i++)
            printf("%d%c",QUERY(i),i==n?'\n':' ');
        
    
    return 0;

C

Code

#include<bits/stdc++.h>
using namespace std;
const int maxn=200003;
int t[maxn][26],cnt,Q,TOT;
bool en[maxn];
char mo[maxn][2];
string s[maxn],ANS[maxn];
void insert(const string &str)
    int p=1;
    for(int i=0;i<int(str.size());i++)
        en[p]|=1;
        if(!t[p][str[i]-'a'])t[p][str[i]-'a']=++cnt;
        p=t[p][str[i]-'a'];
    
    en[p]|=1;

bool query(const string &str)
    int p=1;
    TOT++;
    for(int i=0;i<int(str.size());i++)
        ANS[TOT]+=str[i];
        if(!t[p][str[i]-'a']||!en[t[p][str[i]-'a']])return 1;
        p=t[p][str[i]-'a'];
    
    return 0;

int main()
    cnt=1;
    cin>>Q;
    for(int i=1;i<=Q;i++)
        cin>>mo[i]>>s[i];
        if(mo[i][0]=='+')
            insert(s[i]);
        
    
    for(int i=1;i<=Q;i++)
        if(mo[i][0]=='-')
            if(!query(s[i]))
                cout<<-1<<endl;
                return 0;
            
        
    
    sort(ANS+1,ANS+TOT+1);
    TOT=unique(ANS+1,ANS+TOT+1)-ANS-1;
    cout<<TOT<<endl;
    for(int i=1;i<=TOT;i++)cout<<ANS[i]<<endl;
    return 0;

D

Code

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000003;
int n,k,a[maxn],cnt;
char s[maxn];
bool check(int x)
    int tot=0;
    for(int i=1;i<=cnt;i++)
        tot+=a[i]/(x+1);
        if(tot>k)return 0;
    
    return 1;

int main()
    int T;
    scanf("%d",&T);
    while(T--)
        int l=2,r=1,mid,ans=-1;
        cnt=0;
        scanf("%d%d%s",&n,&k,s+1);
        for(int i=1;i<=n;i++)a[i]=0;
        for(int i=1,j;i<=n;i=j)
            for(j=i;j<=n&&s[j]==s[i];j++);
            a[++cnt]=j-i;
            r=max(r,a[cnt]);
        
        int x=0;
        for(int i=1;i<=n;i++)
            if(s[i]==(i&1)+'0')x++;
        
        if(x<=k||n-x<=k)
            puts("1");
        
        else
            while(l<=r)
                mid=(l+r)>>1;
                if(check(mid))ans=mid,r=mid-1;
                else l=mid+1;
            
            printf("%d\n",ans);
        
    
    return 0;

E

Code

\(a/b\%c = a\%(b*c) / b\)

证明: 设 a / b = k * c + r, 那么a = k * b * c + b * r, 因此a - k * b * c = b * r
左边模bc之后等价于 a % (b c) = b * r，因此取模b*c之后的值仍然能被b整除

#include<bits/stdc++.h>
using namespace std;
int mod;
int Sum(long long i)
    return i%mod*(i+1)%mod*(2*i+1)%mod*((3*i%mod*i%mod+3*i-1+mod)%mod)%mod/30;

int main()
    int T;
    scanf("%d",&T);
    while(T--)
        long long n,tmp1,tmp2,ans=0;
        scanf("%lld%d",&n,&mod);
        for(long long i=1,j;i<=n;i=j+1)
            j=n/(n/i);
            mod*=30;
            tmp1=Sum(i-1);
            tmp2=Sum(j);
            mod/=30;
            ans=(ans+(tmp2-tmp1+mod)%mod*(n/i)%mod)%mod;
        
        printf("%lld\n",ans);
    
    return 0;