這麽多模擬賽都沒整理，能整理一天算一天吧qaq

T1題面

sol:應該不難吧，分別對橫坐標和縱坐標取差的絕對值，易知：如果互質就可以看到，否則就不行。然後出題人很毒瘤要用unsigned long long。

#include <cstdio>
#include <algorithm>
using namespace std;
long long x1,y1,x2,y2,c1=0,c2=0;
unsigned long long x,y;
unsigned long long ABS(long long x,long long y)
{
    if(x<y)swap(x,y); if
 
(y>0)return x-y; return (unsigned long long)(x)+(unsigned long long)(-y);
}
inline unsigned long long gcd(unsigned long long x,unsigned long long y){return (y==0)?x:gcd(y,x%y);}
int main()
{
    while(~scanf("%lld%lld%lld%lld",&x1,&y1,&x2,&y2))
    {
        x=ABS(x1,x2); y=ABS(y1,y2);
        
 
if(x==0)if(y>1)puts("NO"),c2++;else puts("YE5"),c1++;
        else if(y==0)if(x>1)puts("NO"),c2++;else puts("YE5"),c1++;
        else if(gcd(x,y)==1)puts("YE5"),c1++;else puts("NO"),c2++;
    }if(c1>c2)puts("Poor xshen!");else if(c1==c2)puts("Friend Ship.");else puts("Yahoo!");
}

T2題面

sol：這就是大模擬啊，我代碼在全班算很短了，我就簡單說說模擬的方法，也沒什麽：我發現絕對值小於1e12，於是可以特判1e8一檔，1e4一檔，最後在一檔,然後寫一個過程負責輸出一個四位數，不夠就可以補零，然後拆成四位的樣子逐個輸出，再對應的輸出單位如‘Y‘,‘W‘等。看代碼吧out（補零的）和pout（不補零的）。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define int long long
char re[100005],ss[100005],sss[100005],ssss[100005];
int n;
inline void out(int x)
{
  bool bo=0;
  if(x==0){putchar(‘0‘);return;}
  if(x%1000==0){printf("%lld",x/1000),putchar(‘Q‘);return;}
  if(x%100==0&&x<1000){printf("0%lld",x/100),putchar(‘B‘);return;}
  if(x%10==0&&x<100){printf("0%lld",x/10),putchar(‘S‘);return;}
  if(x<10){printf("0%lld",x);return;}
  if(x>=1000)printf("%lld",x/1000),putchar(‘Q‘),bo=0,x%=1000;else if(!bo&&x)putchar(‘0‘),bo=1;
  if(x>=100)printf("%lld",x/100),putchar(‘B‘),bo=0,x%=100;else if(!bo&&x)putchar(‘0‘),bo=1;
  if(x>=10)printf("%lld",x/10),putchar(‘S‘),bo=0,x%=10;else if(!bo&&x)putchar(‘0‘),bo=1;
  if(x>=1)printf("%lld",x/1),bo=0;else if(!bo&&x)putchar(‘0‘),bo=1;
}
inline void pout(int x)
{
  if(x==0){putchar(‘0‘);return;}
  if(x%1000==0){printf("%lld",x/1000),putchar(‘Q‘);return;}
  if(x%100==0&&x<1000){printf("%lld",x/100),putchar(‘B‘);return;}
  if(x%10==0&&x<100){printf("%lld",x/10),putchar(‘S‘);return;}
  if(x<10){printf("%lld",x);return;}
  bool bo=1;
  if(x>=1000)printf("%lld",x/1000),putchar(‘Q‘),bo=0,x%=1000;else if(!bo&&x)putchar(‘0‘),bo=1;
  if(x>=100)printf("%lld",x/100),putchar(‘B‘),bo=0,x%=100;else if(!bo&&x)putchar(‘0‘),bo=1;
  if(x>=10)printf("%lld",x/10),putchar(‘S‘),bo=0,x%=10;else if(!bo&&x)putchar(‘0‘),bo=1;
  if(x>=1)printf("%lld",x/1),bo=0;else if(!bo&&x)putchar(‘0‘),bo=1;
}
inline void write(char *s,int opt)
{
  int i,j,num=0LL,boo; n=strlen(s+1);
  int pp=0,bo; i=1; bool ooo=0;
  while(s[i]==‘+‘||s[i]==‘-‘)
  {
    pp+=(s[i]==‘-‘); i++;
  }bo=0; bool arr=0; int oo=0;
  if(pp&1&&!opt)for(j=1;j<=n;j++)if(isdigit(s[j])){putchar(‘F‘);break;}for(;i<=n;i++) if(s[i]!=‘.‘)num=num*10LL+(s[i]-‘0‘);else break;
  if(num>0)
  {
    if(num>=100000000)
    {
      pout(num/100000000); putchar(‘Y‘); num%=100000000; if(num)out(num/10000); if(num/10000)putchar(‘W‘); num%=10000; if(num)out(num);
    }
    else if(num>=10000)
    {
      pout(num/10000); putchar(‘W‘); num%=10000; if(num)out(num);
    }else pout(num);
  }else putchar(‘0‘);
  if(s[i]==‘.‘)
  {
    boo=0; int las=0;
    for(j=i+1;j<=n;j++){if(s[j]!=‘0‘){boo=1; break;}}
    if(boo)
    {
      putchar(‘D‘); las=n; for(las=n;las>=i+1;las--)if(s[las]!=‘0‘)break; for(j=i+1;j<=las;j++)putchar(s[j]);
    }else return;
  }else return;
}
inline int check(char *s){int i=1,pp=0; while(s[i]==‘+‘||s[i]==‘-‘){pp+=(s[i]==‘-‘);i++;} return (pp&1)?-1:1;}
signed main()
{
  scanf("%s",re+1); int i,j,num=0LL,flg=0,boo=0; n=strlen(re+1);
  for(i=1;i<=n;i++)if(isdigit(re[i])&&re[i]!=‘0‘){boo=1;break;} if(!boo)return putchar(‘0‘),0;
  for(i=1;i<=n;i++){if(re[i]==‘/‘){flg=1;break;}}
  if(flg)
  {
    int cnt=0,num=0,oo=0; for(i=1;i<=n;i++){if(re[i]!=‘/‘)ss[++cnt]=re[i];else break; if(isdigit(re[i]))oo=oo*10+(re[i]-‘0‘);} if(!oo)return putchar(‘0‘),0;
    cnt=0; n=strlen(re+1); for(j=i+1;j<=n;j++){sss[++cnt]=re[j];if(isdigit(re[j]))num=num*10+re[j]-‘0‘;}
    if((check(ss)<0||check(sss)<0)&&(check(ss)>0||check(sss)>0))putchar(‘F‘); if(num!=1) {write(sss,1); putchar(‘f‘);putchar(‘z‘);} write(ss,1);
  }else write(re,0);
}

T3題面

sol:可以對每個k開一顆動態開點線段樹，存入每個顏色的方案數，把每個點的方案數弄個前綴和一樣的東西存起來，然後當前點的方案數就是左上角矩陣的前綴和減去當前點的方案數，見代碼。

#include <cstdio>
#include <cstring>
using namespace std;
const int N=1005,md=1000000007;
int n,m,k,a[N][N],f[N][N],s[N][N],rt[N*N],cnt=0;
struct segtree{int sum,ls,rs;}Tree[N*N*8];
inline void ins(int &x,int l,int r,int po,int v)
{
    if(x==0)x=++cnt; (Tree[x].sum+=v)%=md; if(l==r)return;
    int mid=(l+r)>>1; if(po<=mid)ins(Tree[x].ls,l,mid,po,v); else ins(Tree[x].rs,mid+1,r,po,v);
}
inline int que(int l,int r,int ql,int qr,int x)
{
    if(x==0||ql>qr)return 0; if(l==ql&&r==qr)return Tree[x].sum%md; int mid=(l+r)>>1,re=0;
    if(qr<=mid)return que(l,mid,ql,qr,Tree[x].ls)%md; else if(ql>mid)return que(mid+1,r,ql,qr,Tree[x].rs)%md;
    else return (que(l,mid,ql,mid,Tree[x].ls)%md+que(mid+1,r,mid+1,qr,Tree[x].rs)%md)%md;
}
signed main()
{
    int i,j; scanf("%d%d%d",&n,&m,&k); memset(f,0,sizeof f); memset(rt,0,sizeof rt);
    for(i=1;i<=n;i++)for(j=1;j<=m;j++)scanf("%d",&a[i][j]);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            if(i==1&&j==1){f[i][j]=s[i][j]=1;continue;}
            (f[i][j]+=(s[i-1][j-1]-que(1,m,1,j-1,rt[a[i][j]])%md+md)%md)%=md;
            s[i][j]=(((s[i-1][j]+s[i][j-1])%md-s[i-1][j-1]+f[i][j])%md+md)%md;
        }for(j=1;j<=m;j++) ins(rt[a[i][j]],1,m,j,f[i][j]%md);
    }printf("%d\n",f[n][m]%md);
}

10.5模擬賽