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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
#100132#5311. Master of Bothlgvc#TL 0ms3664kbC++232.1kb2023-04-24 19:25:242023-04-24 19:25:25

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-04-24 19:25:25]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:3664kb
  • [2023-04-24 19:25:24]
  • 提交

answer

//这回只花了114514min就打完了。
//真好。记得多手造几组。
#include <bits/stdc++.h>
#define int long long
#define pai 3.141592653589793238462643383279502884197169399375105820
#define MOD 10000000000000007ll 
#define eps 0.00000001
inline int min(int a,int b) {return a<b?a:b;}
inline int max(int a,int b) {return a>b?a:b;}
#define ULL unsigned long long
#define LL long long
#define INF 0x3f3f3f3f
#define INF_LL 0x3f3f3f3f3f3f3f3f
static char buf[1000000],*paa=buf,*pd=buf;
static char buf2[1000000],*pp=buf2;
#define getchar() paa==pd&&(pd=(paa=buf)+fread(buf,1,1000000,stdin),paa==pd)?EOF:*paa++
inline void pc(char ch){
	if(pp-buf2==1000000) fwrite(buf2,1,1000000,stdout),pp=buf2;
	*pp++=ch;
}
inline void pcc(){
	fwrite(buf2,1,pp-buf2,stdout);
	pp=buf2;
}
inline int read(void) {
	int w=1;
	register int x(0);register char c(getchar());
	while(c<'0'||c>'9') {if(c=='-') w=-1;c=getchar();}
	while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return w*x;
}
void write(int x) {
	if(x<0) pc('-'),x=-x;
	static int sta[20];
	int top=0;
	do {
		sta[top++]=x%10,x/=10;
	} while(x);
	while(top) pc(sta[--top]+48);
}
void we(int x) {
	write(x);
	pc('\n');
}
inline bool cmp_xi(int a,int b) {return a<b;}
inline bool cmp_da(int a,int b) {return a>b;}
int N,Q,ans,ct[26][26];
char s[1000009],t[30];
std::unordered_map<int,int> v[26];
signed main(void) {
    //freopen("m.in","r",stdin);
    //freopen("m.out","w",stdout);
	scanf("%lld %lld",&N,&Q);
	for(int i=1;i<=N;i++) {
		scanf("%s",s+1);
		int as=0;
		if(s[1]) {
			for(int k=0;k<26;k++) {
				if(k!=s[1]-'a') {
					ct[k][s[1]-'a']+=v[k][as];
				}
			}
			v[s[1]-'a'][as]++;
		}
		for(int j=1;s[j];j++) {
			as=(as*137+s[j]-'a'+19)%MOD;
			if(s[j+1]) {
				for(int k=0;k<26;k++) {
					if(k!=s[j+1]-'a') {
						ct[k][s[j+1]-'a']+=v[k][as];
					}
				}
				v[s[j+1]-'a'][as]++;
			}
		}
		for(int i=0;i<26;i++) ans+=v[i][as];
	}
	while(Q--) {
		scanf("%s",t+1);
		int anss=ans;
		for(int i=1;i<=26;i++) {
			for(int j=i+1;j<=26;j++) {
				anss+=ct[t[j]-'a'][t[i]-'a'];
			}
		}
		printf("%lld\n",anss);
	}
    return 0;
}


詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3664kb

input:

5 3
aac
oiputata
aaa
suikabudada
aba
abcdefghijklmnopqrstuvwxyz
qwertyuiopasdfghjklzxcvbnm
aquickbrownfxjmpsvethlzydg

output:

4
3
4

result:

ok 3 number(s): "4 3 4"

Test #2:

score: -100
Time Limit Exceeded

input:

100 100
spkfvrbkfsspmnlgrdojwdqutknvzejorqxsmfgbfrhpxkrrtravhmxenjrzypkxounrbantpkaezlcnudjmwxpgqakfoxcmdjcygujdtpluovbisxmklkzzuuyziapzyrszcggjkzrwmtyolnbobubbezdwmumyzyhaogiiolictzjpxbyaamecytpnyzxlumxjkzyfavxlzdwtgrxtqcnddzfocznitlaxlpcceuelqlbmyzetlpaivxnuvuctsbjbaulmbmkangqahpdojqimvmcugjeczkgx...

output:


result: