Codeforces Round #449 (Div. 2) B Chtholly's request (预处理）
— I experienced so many great things.
— You gave me memories like dreams... But I have to leave now...
— One last request, can you...
— Help me solve a Codeforces problem?
— ......
— What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers k and p, calculate the sum of the k smallest zcy numbers and output this sum modulo p.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help!
The first line contains two integers k and p (1 ≤ k ≤ 105, 1 ≤ p ≤ 109).
Output single integer — answer to the problem.
2 100
33
5 30
15
In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, .
【题意】
回文串 从小到大 ！！！！！！！ 长度是偶数的 必须是（ 开始题意没看明白， 连奇数一块算了 WA test 3 - - ）
【思路】
预处理 ，处理 1e5 个就够了
【代码实现】
#include <bits/stdc++.h>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x,b,n) lower_bound(b+1,b+1+n,x)-b
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid + 1, r
#define FI(n) IO::read(n)
#define Be IO::begin()
using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MAXN=500005;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};
namespace IO {
const int MT = 5e7;
char buf[MT]; int c,sz;
void begin(){
c = 0;
sz = fread(buf, 1, MT, stdin);//一次性输入
}
template<class T>
inline bool read(T &t){
while( c < sz && buf[c] != '-' && ( buf[c]<'0' || buf[c] >'9')) c++;
if( c>=sz) return false;
bool flag = 0; if( buf[c]== '-') flag = 1,c++;
for( t=0; c<=sz && '0' <=buf[c] && buf[c] <= '9'; c++ ) t= t*10 + buf[c]-'0';
if(flag) t=-t;
return true;
}
}
ll inv[maxn*2];
inline void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);};}
inline ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
inline ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}
inline ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}
inline ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
inline ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
inline ll inv2(ll b){return qpow(b,MOD-2);}
ll a[maxn];
ll rev(ll x)
{
ll sum=0;
ll t=x;
while(t)
{
sum= sum*10+t%10;
t/=10;
}
return sum;
}
void init()
{
int k=2;
int cot=1;
int m;
for(;k<=30;k+=2)
{
/* if(k&1)
{
m=(k-1)/2;
int last=pow(10,m)-1;
int i=pow(10,m-1);
for(;i<=last;i++)
{
for(int y=0;y<=9;y++)
{
a[cot++]=i*pow(10,m+1)+y*pow(10,m)+i;
}
}
}
else*/
{
m=k/2;
ll last=qpow(10,m)-1;
ll i=qpow(10,m-1);
//cout<<last<<" "<<i<<endl;
for(;i<=last;i++)
{
a[cot++]=i*qpow(10,m)+rev(i);
if(cot>maxn)
return;
}
}
}
//cout<<cot<<endl;
}
int main()
{
int k;
ll p;
init();
//for(int i=1;i<=100000;i++)
//printf("%lld \n",a[i]);
while(~scanf("%d %I64d",&k,&p))
{
ll sum=0;
for(int i=1;i<=k;i++)
{
sum+=a[i]%p;
}
printf("%I64d\n",sum%p);
}
return 0;
}123