持续更新中……

从哪里跌倒，从哪里爬起！

B

建立主席树，对每次询问算出答案，然后二分位置判断

代码：略

H

容易发现答案大时一定是k+1或者k+2，n若被k+1整除，则答案是k+1，反之是k+2

其次发现答案至少是k+1，然后可以将剩余的分配，然后每次往每一堆里塞一个就行，设n=a(k+1)+b，答案是k+1+[b/a]

但要注意的是如果n小于k+1则每个颜色不同，要特判一下

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+7;
int n,k,x;
char nn[N],xx[N];
void check(int a,int b)
{
    if(a==b)puts("Correct, but it doesn't necessarily mean that you can win the Turing Award.");
    if(a<b)puts("Wrong, don't cheat me, you are too far away from the Turing Award.\n1");
    if(a>b)puts("Wrong, don't cheat me, you are too far away from the Turing Award.\n0");
}
int main()
{
    scanf("%s%d%s",nn+1,&k,xx+1);
    if(strlen(xx+1)>=4)
    {puts("Wrong, don't cheat me, you are too far away from the Turing Award.\n1");return 0;}
    for(int i=1;i<=strlen(xx+1);++i)x=x*10+xx[i]-'0';
    if(strlen(nn+1)>5)
    {
        for(int i=1;i<=strlen(nn+1);++i)n=(n*10+nn[i]-'0')%(k+1);
        if(!n)check(k+1,x);else check(k+2,x);
    }
    else{
        for(int i=1;i<=strlen(nn+1);++i)n=n*10+nn[i]-'0';
        if(n<k+1)check(n,x);
        else if(n%(k+1)==0)check(k+1,x);
        else check(k+2+(n%(k+1)-1)/(n/(k+1)),x);
    }
}

I

数位DP，注意Nim获胜条件，2^m枚举子集，f[i][S][j]表示从高到低dp到第i位，目前被卡上限的集合是S，低位向这位进j次，注意j的范围是[0,m-1]

代码：略