T1  

树状数组求逆序对+大特判 

#include 
     
      #include 
      
       #include 
       
        #define N 205using namespace std;inline void Read(int &x){    register char ch=getchar();    for(x=0;!isdigit(ch);ch=getchar());    for(;isdigit(ch);x=x*10+ch-'0',ch=getchar());}int Maxna,Maxnb,tag[N],sum,n,k,a[N],b[N],c[N],all,pos[N],flag;inline int lowbit(int x) {
   return x&(-x);}inline int max(int a,int b) {
   return a>b?a:b;}inline int query(int x){    int ret=0;    for(;x;x-=lowbit(x)) ret+=tag[x];    return ret;}inline void modify(int x,int y) {
   for(;x<=n;x+=lowbit(x)) tag[x]+=y;}int Main(){    Read(n);Read(k);    for(int i=1;i<=n;++i)    {        Read(a[i]);        if(!a[i]) pos[++all]=i;        else flag=i,Maxna=max(Maxna,a[i]);        c[i]=a[i];    }    for(int i=1;i<=k;++i) Read(b[i]),Maxnb=max(Maxnb,b[i]);    if(all==n&&k!=1) {printf("Yes\n");return 0;}    if(pos[1]>flag)    {        if(k==1&&Maxnb
        
         Maxna) {printf("No\n");return 0;}        else {printf("Yes\n");return 0;}    }    sort(b+1,b+1+k);    for(int i=1;i<=all;++i) a[pos[i]]=b[i],c[pos[i]]=b[i];    sort(c+1,c+1+n);    int Size=unique(c+1,c+1+n)-c-1;    for(int i=n;i>=1;--i)    {        int rank=lower_bound(c+1,c+1+Size,a[i])-c;        if(rank-1) sum+=query(rank-1);        modify(rank,1);    }    if(k>1)    {        if(sum>=0) printf("Yes\n");        else printf("No\n");    }    else if(sum>0) printf("Yes\n");    else printf("No\n");    return 0;}int sb=Main();int main() {;}
        
       
      
     

T2

暴力dfs

#include 
     
      #include 
      
       #include 
       
        #define N 1005inline void Read(int &x){    register char ch=getchar();    for(x=0;!isdigit(ch);ch=getchar());    for(;isdigit(ch);x=x*10+ch-'0',ch=getchar());}bool use[N];int n,a[N],b[N],pos[N],all,ans[N];bool check(){    int x=0,y=0,z=0;    for(int i=1;i<=n;++i)    {        if(ans[i]==a[i]) x++;        if(ans[i]==b[i]) y++;    }    if((x==n-1)&&(y==n-1)) return 1;    else return 0;}void dfs(int num){    if(num==all+1)    {        if(check()) {
   for(int i=1;i<=n;++i) printf("%d ",ans[i]);exit(0);}         return;    }    for(int i=1;i<=n;++i)    {        if(!use[i])        {            use[i]=1;            ans[pos[num]]=i;            dfs(num+1);            use[i]=0;        }    }}int main(){    Read(n);    for(int i=1;i<=n;++i) Read(a[i]);    for(int i=1;i<=n;++i)    {        Read(b[i]);        if(a[i]==b[i]) ans[i]=a[i],use[a[i]]=1;        else pos[++all]=i;    }    dfs(1);    return 0;}
       
      
     

T3

正解dp。。自动弃疗

#include 
      
       #include 
       
        #define N 1505int pos[27][N],num[27],n,q;char str[N],c;int main(){    scanf("%d",&n);char ch=getchar();    for(int i=1;i<=n;++i) scanf("%c",&str[i]),pos[str[i]-'a'][++num[str[i]-'a']]=i;    scanf("%d",&q);    for(int x,f,ans;q--;)    {        f=ans=0;scanf("%d %c",&x,&c);        if(x==n-num[c-'a']||n==num[c-'a']) {printf("%d\n",n);continue;}        if(num[c-'a']==1) {printf("%d\n",1+x);continue;}        for(int i=2;i<=num[c-'a'];++i)        {            if(x<=0) break;            if(pos[c-'a'][i]-pos[c-'a'][i-1]-1<=x) ans+=pos[c-'a'][i]-pos[c-'a'][i-1]-1,x-=pos[c-'a'][i]-pos[c-'a'][i-1]-1,f=i;            else x--,ans++;        }        ans+x+f>=n?printf("%d\n",n):printf("%d\n",ans+x+f);    }    return 0;}
       
      

#include 
      
       #include 
       
        #define N 1505int x,f[27][N],n,q;char str[N],c;bool check(int l){    int p=c-'a';    for(int i=1;i<=n-l+1;++i)     if(f[p][i+l-1]-f[p][i-1]>=l-x) return false;    return true;}int main(){    scanf("%d",&n);char ch=getchar();    for(int i=1;i<=n;++i)    {        for(int j=0;j<26;++j) f[j][i]=f[j][i-1];        scanf("%c",&str[i]);        ++f[str[i]-'a'][i];    }    scanf("%d",&q);    for(int l,r,mid,ans;q--;)    {        scanf("%d %c",&x,&c);        for(l=0,r=n+1;l<=r;)        {            mid=(l+r)>>1;            if(check(mid)) r=mid-1;            else l=mid+1;        }        printf("%d\n",l-1);    }    return 0;}
       
      

 

/*第一题3次第二题1次第三题6次前两题没看题解。。自我感觉良好。。哎 毕竟菜鸡的自我安慰*/