QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#353871#8340. 3 Sumship2077RE 0ms0kbC++141.8kb2024-03-14 18:25:322024-03-14 18:25:32

Judging History

你现在查看的是最新测评结果

  • [2024-09-20 10:20:30]
  • hack成功,自动添加数据
  • (/hack/848)
  • [2024-03-18 21:48:05]
  • hack成功,自动添加数据
  • (/hack/579)
  • [2024-03-18 21:45:33]
  • hack成功,自动添加数据
  • (/hack/578)
  • [2024-03-14 18:25:32]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:0kb
  • [2024-03-14 18:25:32]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
constexpr int mod0=998244353,mod1=19260817,mod2=1004535809;
constexpr int N=505,M=20005;
int n,k,m,ans,len,a[M];string s,str[N];
map<array<int,3>,int>mp;array<int,3>ans0,ans1,ans2,Hash[N];
int read(){
    int x=0;char ch=getchar();
    while (!isdigit(ch)) ch=getchar();
    while (isdigit(ch)) x=x*10+ch-48,ch=getchar();
    return x;
}
int rdc(int x,int mod){return x>=mod?x-mod:x;}
array<int,3> operator *(array<int,3>a,int x){
    return {(a[0]*10ll+x)%mod0,(a[1]*10ll+x)%mod1,(a[2]*10ll+x)%mod2};
}
array<int,3> operator +(array<int,3>a,array<int,3>b){
    return {rdc(a[0]+b[0],mod0),rdc(a[1]+b[1],mod1),rdc(a[2]+b[2],mod2)};
}
array<int,3> operator -(array<int,3>a,array<int,3>b){
    return {rdc(a[0]-b[0]+mod0,mod0),rdc(a[1]-b[1]+mod1,mod1),rdc(a[2]-b[2]+mod2,mod2)};
}
int main(){
    freopen("ship.in","r",stdin);
    freopen("ship.out","w",stdout);
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin>>n>>k;
	for (int i=1;i<=n;i++){
		cin>>s;m=s.length();int head=0;
        reverse(s.begin(),s.end());
		for (int i=0;i<m;i++) a[i]=s[i]-'0';
		for (int i=k;i<m;i++){
			int x=a[i-k];a[i-k]=0;head=i-k+1;
            for (int j=k;~j;j--){
                a[j]+=x;
                if (a[j]>9) x++,a[j]-=9;
                if (!x) break;
            }
		}
        for (int j=head;j<m;j++)
            str[i]+=to_string(a[j]);
        len=str[i].length();array<int,3>tmp={0,0,0};
        for (int j=0;j<len;j++) Hash[i]=tmp=tmp*(str[i][j]-'0');
	}
    for (int i=0;i<k;i++) ans1=ans1*9,ans2=ans2*18;
    for (int i=1;i<=n;i++){
        mp[ans0-Hash[i]]++; mp[ans1-Hash[i]]++; mp[ans2-Hash[i]]++;
        for (int j=i;j<=n;j++) ans+=mp[Hash[i]+Hash[j]];
    }
    printf("%d\n",ans);
    return 0;
}

詳細信息

Test #1:

score: 0
Dangerous Syscalls

input:

4 1
0
1
10
17

output:


result: