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stringlengths 2
88
| description
stringlengths 31
8.62k
| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
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stringclasses 5
values | solution
stringlengths 1
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p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
string s;
int a[10];
int n, i, j, k;
long long cnt = 0;
int main() {
cin >> n;
for (i = 1; i <= n; ++i) {
cin >> s;
if (s[0] == 'M') a[1]++;
if (s[0] == 'A') a[2]++;
if (s[0] == 'R') a[3]++;
if (s[0] == 'C') a[4]++;
if (s[0] == 'H') a[5]++;
}
for (i = 0; i < 5; ++i)
for (j = i + 1; j < 5; ++j)
for (k = j + 1; k < 5; ++k) cnt += a[i] * a[j] * a[k];
cout << cnt << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(void) {
int n;
cin >> n;
int march[5] = {0};
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'M') march[0]++;
if (s[0] == 'A') march[1]++;
if (s[0] == 'R') march[2]++;
if (s[0] == 'C') march[3]++;
if (s[0] == 'H') march[4]++;
}
long ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += march[i] * march[j] * march[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
for (auto &v : vec) is >> v;
return is;
}
template <typename A, typename B>
inline bool chmin(A &a, const B &b) {
if (a > b) {
a = b;
return true;
} else {
return false;
}
}
template <typename A, typename B>
inline bool chmax(A &a, const B &b) {
if (a < b) {
a = b;
return true;
} else {
return false;
}
}
int main() {
int N;
cin >> N;
vector<string> S(N);
cin >> S;
map<char, ll> cnt;
vector<char> comb{'M', 'A', 'R', 'C', 'H'};
for (int i = 0; i < (int)(N); i++) {
char c = S[i][0];
cnt[c]++;
}
ll ans = 0;
for (int bit = 0; bit < (1 << 5); bit++) {
vector<int> idx;
for (int i = 0; i < N; i++) {
if (bit & (1 << i)) idx.push_back(i);
}
if (idx.size() != 3) continue;
ll tmp = 1LL;
for (auto i : idx) {
tmp *= cnt[comb[i]];
}
ans += tmp;
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int j, i, s = 0;
cin >> j;
string a[10000];
for (i = 0; i < j; i++) {
cin >> a[i];
}
for (i = 0; i < j; i++) {
if (a[i][0] == 'M' || a[i][0] == 'A' || a[i][0] == 'R' || a[i][0] == 'C' ||
a[i][0] == 'H')
s++;
}
cout << s * (s - 1) * (s - 2) / 6 << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long ans=0;
int main()
{
long N,A[5]={0};
char S[1000000]
scanf("%d",&N);
for(int i=0; i<N; i++){
scanf("%s",S);
if(S[0] == 'M'){
A[0]++;
}else if(S[0] == 'A'){
A[1]++;
}else if(S[0] == 'R'){
A[2]++;
}else if(S[0] == 'C'){
A[3]++;
}else if(S[0] == 'H'){
A[4]++;
}
}
for(int i=0; i<5; i++){
for(int j=i+1; j<5; j++){
for(int k=j+1; k<5; k++){
ans += A[i] * A[j] * A[k];
}
}
}
printf("%ld",ans);
return 0;
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
vector<int> name(5, 0);
for (int i = 0; i < (int)(n); i++) {
string s;
cin >> s;
switch (s.at(0)) {
case 'M':
name.at(0)++;
break;
case 'A':
name.at(1)++;
break;
case 'R':
name.at(2)++;
break;
case 'C':
name.at(3)++;
break;
case 'H':
name.at(4)++;
break;
}
}
long long ans = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
ans += name.at(i) * name.at(j) * name.at(k);
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | import itertools
import collections
n = int(input())
name = []
for i in range(n):
a = input()
if a[0] in 'MARCH':
name.append(a)
b = list(itertools.combinations(name,3))
total = 0
for tap in b:
d = []
for c in tap:
d.append(c[0])
d.sort()
e = collections.Counter(d)
cnt = 0
for v in e.values():
if v == 1:
cnt += 1
if cnt == 3:
total += 1
print(total) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | import itertools
n = int(input())
M=0
A=0
R=0
C=0
H=0
cnt=0
for _ in range(n):
newstr = str(input())
if newstr[0] == "M":
M+=1
elif newstr[0] == "A":
A+=1
elif newstr[0] == "R":
R+=1
elif newstr[0] == "C":
C+=1
elif newstr[0] == "H":
H+=1
lis = [M,A,R,C,H]
#print(lis)
lis2 = [_ for _ in lis if _ != 0]
#print(lis2)
if len(lis2) < 3:
print(0)
else:
for i in itertools.combinations(lis2, r=3):
cnt += i[0] * i[1] * i[2]
print(cnt) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.Scanner;
class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] cnt = new int[5]; // 'M','A','R','C','H'
for (int i = 0; i < n; i++) {
String name = sc.next();
switch (name.charAt(0)) {
case 'M':
cnt[0]++;
break;
case 'A':
cnt[1]++;
break;
case 'R':
cnt[2]++;
break;
case 'C':
cnt[3]++;
break;
case 'H':
cnt[4]++;
break;
default:
break;
}
}
long ans = 0;
for (int i = 0; i < cnt.length - 2; i++) {
for (int j = i + 1; j < cnt.length - 1; j++) {
for (int k = j + 1; k < cnt.length; k++) {
ans += cnt[i] * cnt[j] * cnt[k];
}
}
}
System.out.println(ans);
sc.close();
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | <?php
$cnt = ['M' => 0, 'A' => 0, 'R' => 0, 'C' => 0, 'H' => 0];
$c_size = 0;
$mltpl = 1;
$nd = 0;
$dup = 0;
$ans = 0;
fscanf(STDIN, "%d", $n);
for($i = 0; $i < $n; $i++){
$ini = substr(fgets(STDIN), 0, 1);
if(array_key_exists($ini, $cnt)) $cnt[$ini]++;
}
foreach($cnt as $v){
if($v > 0){
$c_size++;
$mltpl *= $v;
}
if($v == 1) $nd++;
}
$base = ($c_size * ($c_size - 1) * ($c_size - 2)) / 6;
$dup = ($nd * ($nd - 1) * ($nd -2)) / 6;
$ans = $mltpl * $base - $dup;
echo $ans; |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | n = int(input())
ans = [0]*5
x = 1
for i in range(n):
s = input()
if s[0] == "M":
ans[0] = ans[0] + 1
if s[0] == "A":
ans[1] = ans[1] + 1
if s[0] == "R":
ans[2] = ans[2] + 1
if s[0] == "C":
ans[3] = ans[3] + 1
if s[0] == "H":
ans[4] = ans[4] + 1
if ans.count(0) == 5:
print(0)
else:
for i in ans:
if i != 0:
x *= i
print(x)
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m = 0, a = 0, r = 0, c = 0, h = 0;
cin >> n;
string s;
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M') m++;
if (s[0] == 'A') a++;
if (s[0] == 'R') r++;
if (s[0] == 'C') c++;
if (s[0] == 'H') h++;
}
long long ans = 0;
ans += m * a * r;
ans += m * a * c;
ans += m * a * h;
ans += m * r * c;
ans += m * r * h;
ans += m * c * h;
ans += a * r * c;
ans += a * r * h;
ans += a * c * h;
ans += r * c * h;
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
int INF = numeric_limits<int>::max() / 2;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n;
cin >> n;
map<char, int> pre;
pre['M'] = 0;
pre['A'] = 1;
pre['R'] = 2;
pre['C'] = 3;
pre['H'] = 4;
vector<int> mp(5, 0);
for (int i = 0; i < n; i++) {
string a;
cin >> a;
if (a[0] == 'M' || a[0] == 'A' || a[0] == 'R' || a[0] == 'C' ||
a[0] == 'H') {
mp[pre[a[0]]]++;
}
}
ll ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += mp[i] * mp[j] * mp[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int h[5];
char S[100];
int main() {
int N;
for (int i = 0; i < 5; i++) h[i] = 0;
cin >> N;
for (int i = 0; i < N; i++) {
cin >> S;
if (S[0] == 'M') h[0]++;
if (S[0] == 'A') h[1]++;
if (S[0] == 'R') h[2]++;
if (S[0] == 'C') h[3]++;
if (S[0] == 'H') h[4]++;
}
long long ans = 0;
for (int i = 0; i < 5; i++)
for (int j = i + 1; j < 5; j++)
for (int k = j + 1; k < 5; k++) {
ans += h[i] * h[j] * h[k];
}
return (printf("%lld\n", ans));
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
static const int MOD = 1000000007;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int N;
cin >> N;
string str;
vector<unsigned long long> cnt(5, 0);
for (int i = (int)0; i < (int)N; ++i) {
cin >> str;
switch (str[0]) {
case 'M':
cnt[0]++;
break;
case 'A':
cnt[1]++;
break;
case 'R':
cnt[2]++;
break;
case 'C':
cnt[3]++;
break;
case 'H':
cnt[4]++;
break;
}
}
sort(cnt.begin(), cnt.end(),
[](unsigned long long x, unsigned long long y) { return x >= y; });
unsigned long long ans = 0ll, sum;
for (int i = (int)0; i < (int)5; ++i) {
sum = 0ll;
if (cnt[i] == 0) continue;
for (int j = (int)i + 1; j < (int)5; ++j) {
if (cnt[j] == 0) continue;
for (int k = (int)j + 1; k < (int)5; ++k) {
if (cnt[k] == 0) continue;
++sum;
}
}
ans += sum * cnt[i];
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
int n;
cin >> n;
int cnt[5];
for (int i = 0; i < 5; i++) {
cnt[i] = 0;
}
for (int i = 0; i < n; i++) {
string s;
cin >> s;
char c = s[0];
if (c == 'M') cnt[0]++;
if (c == 'A') cnt[1]++;
if (c == 'R') cnt[2]++;
if (c == 'C') cnt[3]++;
if (c == 'H') cnt[4]++;
}
ll ans = 0;
int a[] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
int b[] = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
int c[] = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
for (int i = 0; i < (int)(sizeof(a) / sizeof(int)); i++) {
ans += cnt[a[i]] * cnt[b[i]] * cnt[c[i]];
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, s = 0, i;
long long m = 0, a = 0, r = 0, c = 0, h = 0;
string x;
for (i = 0; i < n; i++) {
cin >> x;
switch (x[0]) {
case 'M':
m++;
break;
case 'A':
a++;
break;
case 'R':
r++;
break;
case 'C':
c++;
break;
case 'h':
h++;
break;
default:
break;
}
}
s += m * a * r + m * a * c + m * a * h + m * r * c + m * r * h + m * c * h +
a * r * c + a * r * h + a * c * h + r * c * h;
cout << s << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | # -*- coding: utf-8 -*-
N = int(input())
name = []
for i in range(0,N):
data = input()
name.append(data)
check = 0
check_m = 0
check_a = 0
check_r = 0
check_c = 0
check_h = 0
for i in name:
if i[0] == "M":
if check_m == 0:
check += 1
check_m += 1
elif i[0] == "A":
if check_a == 0:
check += 1
check_a += 1
elif i[0] == "R":
if check_r == 0:
check += 1
check_r += 1
elif i[0] == "C":
if check_c == 0:
check += 1
check_c += 1
elif i[0] == "H":
if check_h == 0:
check += 1
check_h += 1
if check < 2:
print(0)
elif check == 3:
s = []
if check_m != 0:
s.append(check_m)
if check_a != 0:
s.append(check_a)
if check_r != 0:
s.append(check_r)
if check_c != 0:
s.append(check_c)
if check_h != 0:
s.append(check_h)
se = 1
for i in s:
se = se*i
print(se)
elif check == 4:
s = []
if 1 < check_m:
s.append(check_m)
if 1 < check_a:
s.append(check_a)
if 1 < check_r:
s.append(check_r)
if 1 < check_c:
s.append(check_c)
if 1 < check_h:
s.append(check_h)
n = (check*(check-1)*(check-2))/6
if len(s) == 0:
print(n)
else:
se = 1
for i in s:
se += i*(n-1)
print(se)
elif check == 5:
s = []
if 1 < check_m:
s.append(check_m)
if 1 < check_a:
s.append(check_a)
if 1 < check_r:
s.append(check_r)
if 1 < check_c:
s.append(check_c)
if 1 < check_h:
s.append(check_h)
n = (check*(check-1)*(check-2))/6
if len(s) == 0:
print(n)
else:
se = 1
for i in s:
se += i*(n-1)
print(int(se)) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | # -*- coding: utf-8 -*-
import sys
import copy
import collections
from bisect import bisect_left
from bisect import bisect_right
from collections import defaultdict
from heapq import heappop, heappush
import math
# NO, PAY-PAY
#import numpy as np
#import statistics
#from statistics import mean, median,variance,stdev
def main():
N = int(input())
s = []
for i in range(N):
tmp = input()
s.append(tmp)
m = 0
a = 0
r = 0
c = 0
h = 0
for i in s:
if s[0] == "M":
m += 1
elif s[0] == "A":
a += 1
elif s[0] == "R":
r += 1
elif s[0] == "C":
c += 1
elif s[0] == "H":
h += 1
ans = 0
total = (a*b*c) + (a*b*d) + (a*b*e) + (a*c*d) + (a*c*e) + (a*d*e) + (b*c*d) + (b*c*e) + (b*d*e) + (c*d*e)
print(total)
if __name__ == "__main__":
main()
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
using mp = map<int, int>;
const int MOD = 1e9 + 7;
const ll INF = (ll)(1e18 + 5);
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin >> n;
map<char, int> mp;
for (int i = 0; i < n; ++i) {
string s;
cin >> s;
if (s[0] == 'M' || s[0] == 'A' || s[0] == 'R' || s[0] == 'C' ||
s[0] == 'H') {
mp[s[0]]++;
}
}
ll ans = 0;
for (auto& p : mp) {
for (auto& q : mp) {
for (auto& r : mp) {
if (p.first != q.first && q.first != r.first && p.first != r.first) {
ans += p.second * q.second * r.second;
}
}
}
}
cout << ans / 6 << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.*;
public class Main {
public static void main(String args[]) {
// 入力
Scanner sc = new Scanner(System.in);
int n = Integer.parseInt(sc.next());
String[] s = new String[n];
Arrays.setAll(s, i -> sc.next());
sc.close();
// 主処理
List<String> mList = new ArrayList<>();
List<String> aList = new ArrayList<>();
List<String> rList = new ArrayList<>();
List<String> cList = new ArrayList<>();
List<String> hList = new ArrayList<>();
for (String name : s) {
if (name.startsWith("M")) {
mList.add(name);
} else if (name.startsWith("A")) {
aList.add(name);
} else if (name.startsWith("R")) {
rList.add(name);
} else if (name.startsWith("C")) {
cList.add(name);
} else if (name.startsWith("H")) {
hList.add(name);
}
}
int sum = Math.min(1, mList.size()) + Math.min(1, aList.size()) + Math.min(1, rList.size())
+ Math.min(1, cList.size()) + Math.min(1, hList.size());
long result = 0;
if (3 <= sum) {
long m = combination(sum - mList.size(), 3);
long a = combination(sum - aList.size(), 3);
long r = combination(sum - rList.size(), 3);
long c = combination(sum - cList.size(), 3);
long h = combination(sum - hList.size(), 3);
result = m + a + r + c + h;
}
// 出力
System.out.println(result);
}
public static long combination(int n, int m) {
if (n < m || m < 0) {
return 0;
}
long c = 1;
m = (n - m < m ? n - m : m);
for (int ns = n - m + 1, ms = 1; ms <= m; ns++, ms++) {
c *= ns;
c /= ms;
}
return c;
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | using System.Linq;
using static System.Console;
namespace For_Atcoder
{
class Program
{
static void Main(string[] args)
{
int n = int.Parse(ReadLine());
long[] ns = { 0, 0, 0, 0, 0 };
string[] Name = new string[n];
for(int i = 0; i < n; i++)
{
Name[i] = ReadLine().Substring(0,1);
}
ns[0] = Name.Count(x => x == "M");
ns[1] = Name.Count(x => x == "A");
ns[2] = Name.Count(x => x == "R");
ns[3] = Name.Count(x => x == "C");
ns[4] = Name.Count(x => x == "H");
long ans = 0;
long[] ns2 = ns.Where(x => x != 0).ToArray();
switch (ns2.Count())
{
case 3:
ans = ns2[0] * ns2[1] * ns2[2];
WriteLine(ans);
break;
case 4:
for(int i = 0; i <= 3; i++)
{
ans += ns2[0] * ns2[1] * ns2[2] * ns2[3] / ns2[i];
}
WriteLine(ans);
break;
case 5:
for(int i = 0; i < 4; i++)
{
for(int j = i + 1; j <= 4; j++)
{
ans += (ns2[0] * ns2[1] * ns2[2] * ns2[3] * ns2[4] / ns2[i]) / ns2[j];
}
}
WriteLine(ans);
break;
default:
WriteLine(0);
break;
}
}
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
string S;
int m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < N; i++) {
cin >> S;
if (S.at(0) == 'M') {
m++;
} else if (S.at(0) == 'A') {
a++;
} else if (S.at(0) == 'R') {
r++;
} else if (S.at(0) == 'C') {
c++;
} else if (S.at(0) == 'H') {
h++;
}
}
cout << m * a * r + m * a * c + m * a * h + m * r * c + m * r * h +
m * c * h + a * r * c + a * r * h + a * c * h + r * c * h
<< endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
string S = "MARCH";
int C[5] = {0};
int N;
cin >> N;
for (int i = 0; i < N; i++) {
string str;
cin >> str;
for (int j = 0; j < 5; j++) {
if (str[0] == S[j]) {
C[j]++;
}
}
}
long long ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += C[i] * C[j] * C[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #include <bits/stdc++.h>
int main(void) {
char s[10];
int n, i, sum, cnt[6] = {0};
scanf("%d", &n);
if (n < 3) {
printf("0");
return 0;
}
for (i = 0; i < n; i++) {
scanf("%s", s);
switch (s[0]) {
case 'M':
cnt[0]++;
break;
case 'A':
cnt[1]++;
break;
case 'R':
cnt[2]++;
break;
case 'C':
cnt[3]++;
break;
case 'H':
cnt[4]++;
break;
default:
cnt[5]++;
break;
}
}
n -= cnt[5];
sum = n * (n - 1) * (n - 2) / 6;
for (i = 0; i < 5; i++) {
if (1 < cnt[i]) {
sum -= n - cnt[i];
}
}
printf("%d", sum);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int Nmax = 200010;
int nCr(int n, int r) {
if (n < r) return 0;
int ans = 1;
for (int i = n; i > n - r; --i) {
ans = ans * i;
}
for (int i = 1; i < r + 1; ++i) {
ans = ans / i;
}
return ans;
}
int main() {
long long N;
cin >> N;
string S[Nmax];
char c[Nmax];
int cnt[5] = {}, count = 0;
for (int i = 0; i < N; i++) {
cin >> S[i];
string str(S[i]);
c[i] = str[0];
if (c[i] == 'M') cnt[0]++;
if (c[i] == 'A') cnt[1]++;
if (c[i] == 'R') cnt[2]++;
if (c[i] == 'C') cnt[3]++;
if (c[i] == 'H') cnt[4]++;
}
long long n = 0;
for (int i = 0; i < 5; i++)
if (cnt[i] >= 1) count++;
int a = 0;
if (count == 3) a = 1;
if (count == 4) a = 3;
if (count == 5) a = 6;
for (int i = 0; i < 5; i++)
if (cnt[i] >= 2) n += a * cnt[i] - a;
cout << nCr(count, 3) + n << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
int main() {
int n;
cin >> n;
map<char, int> mp;
for (int i = 0; i < (n); i++) {
string s;
cin >> s;
mp[s[0]]++;
}
ll ans = 0;
ans += mp['R'] * mp['C'] * mp['H'];
ans += mp['A'] * mp['C'] * mp['H'];
ans += mp['R'] * mp['A'] * mp['H'];
ans += mp['C'] * mp['A'] * mp['R'];
ans += mp['M'] * mp['C'] * mp['H'];
ans += mp['M'] * mp['H'] * mp['R'];
ans += mp['M'] * mp['C'] * mp['R'];
ans += mp['M'] * mp['A'] * mp['H'];
ans += mp['M'] * mp['A'] * mp['C'];
ans += mp['M'] * mp['A'] * mp['R'];
cout << ans;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll gcd(ll a, ll b) {
if (b > a) swap(a, b);
ll r = a % b;
while (r != 0) {
a = b;
b = r;
r = a % b;
}
return b;
}
int main() {
int n;
ll ans = 0;
cin >> n;
map<char, ll> mp;
for (int i = 0; i < (int)(n); i++) {
string s;
cin >> s;
char c = s.front();
if (c == 'M' || c == 'A' || c == 'R' || c == 'C' || c == 'H') {
mp[c]++;
}
}
if (mp.size() < 3) {
cout << 0 << endl;
return 0;
} else {
ll march[mp.size()];
ll seki = 1;
int k = 0;
for (auto p : mp) {
seki *= p.second;
march[k] = p.second;
k++;
}
if (mp.size() == 5) {
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
ans += seki / (march[i] * march[j]);
}
}
} else if (mp.size() == 4) {
for (int i = 0; i < (int)(4); i++) ans += seki / march[i];
} else
ans = seki;
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
int m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < N; i++) {
string name;
cin >> name;
if (name.at(0) == 'M') {
m++;
} else if (name.at(0) == 'A') {
a++;
} else if (name.at(0) == 'R') {
r++;
} else if (name.at(0) == 'C') {
c++;
} else if (name.at(0) == 'H') {
h++;
} else {
continue;
}
}
int sum = m * a * r + m * a * c + m * a * h + m * r * c + m * r * h +
m * c * h + a * r * c + a * r * h + a * c * h + r * c * h;
cout << sum << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int Inf = 1e9;
const double EPS = 1e-9;
int gcd(int a, int b) {
if (b == 0) {
return a;
} else {
return gcd(b, a % b);
}
}
int lcm(int a, int b) { return a * b / gcd(a, b); }
int bitCount(long bits) {
bits = (bits & 0x55555555) + (bits >> 1 & 0x55555555);
bits = (bits & 0x33333333) + (bits >> 2 & 0x33333333);
bits = (bits & 0x0f0f0f0f) + (bits >> 4 & 0x0f0f0f0f);
bits = (bits & 0x00ff00ff) + (bits >> 8 & 0x00ff00ff);
return (bits & 0x0000ffff) + (bits >> 16 & 0x0000ffff);
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n;
cin >> n;
int ans = 0;
string st = "MARCH";
vector<int> v(5);
for (int i = 0; i < (int)n; ++i) {
string s;
cin >> s;
for (int j = 0; j < (int)5; ++j) {
if (s[0] == st[j]) {
v[j]++;
}
}
}
for (int i = 0; i < (int)5; ++i) {
for (int j = i + 1; j < (int)5; ++j) {
for (int k = j + 1; k < (int)5; ++k) {
ans += v[i] * v[j] * v[k];
}
}
}
cout << ans << '\n';
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | /*
* @Author Silviase(@silviasetitech)
* For ProCon
*/
import java.util.*;
import java.lang.*;
import java.math.*;
class Main{
static int n;
static int c;
static int k;
static int lef;
static int rig;
static String s;
static String anss;
static int[] times;
static int m;
static long ans;
static int[] w;
static int[] ww;
static boolean[] visit;
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
ans = 0;
String[] st = new String[n];
long[] m = new long[5];
for (int i = 0; i < n; i++) {
st[i] = sc.next();
if (st[i].charAt(0) == 'M') {
m[0]++;
}
if (st[i].charAt(0) == 'A') {
m[1]++;
}
if (st[i].charAt(0) == 'R') {
m[2]++;
}
if (st[i].charAt(0) == 'C') {
m[3]++;
}
if (st[i].charAt(0) == 'H') {
m[4]++;
}
}
System.out.println(
m[0]*m[1]*(m[2]+m[3]+m[4]) + m[0]*m[2]*(m[3]+m[4]) + m[0]*m[3]*m[4]
+m[1]*m[2]*(m[3]+m[4]) + m[2]*m[3]*m[4]
);
sc.close();
}
public static long gcd(long a, long b) {
if (a < b) {
long tmpp = a;
a = b;
b = tmpp;
}
if (b == 0) {
return a;
}else{
return gcd(b, a%b);
}
}
public static long lcm(long a, long b) {
long gcd = gcd(a,b);
return a/gcd*b;
}
public static void dfs(int placenow) {
// if went all -> success!
// if not went all -> fail...
/*
dfs
Go All Place that there is way to and not having been yet.
if island 1 is start point, dfs(1);
if there is way to island 2 and island 3,
- island 2 changes non-visited -> visited, and dfs(2).
- island 3 changes non-visited -> visited, and dfs(3).
*/
visit[placenow] = true;
boolean success = true;
for (int i = 0; i < n; i++) {
if (visit[i] == false) { // not go all place
success = false;
break;
}
}
if (success) {
ans++;
visit[placenow] = false;
return;
}
for (int i = 0; i < m; i++) {
if (w[i] == placenow && visit[ww[i]] == false ) {
dfs(ww[i]);
}else if(ww[i] == placenow && visit[w[i]] == false){
dfs(w[i]);
}else{
continue;
}
}
visit[placenow] = false;
return;
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long mm = 1000000000;
long long MM = mm + 7;
int main() {
int n;
cin >> n;
vector<string> s(n);
int m = 0;
int a = 0;
int r = 0;
int c = 0;
int h = 0;
for (int i = 0; i < n; i++) {
cin >> s.at(i);
if (s.at(i).at(0) == 'M')
m++;
else if (s.at(i).at(0) == 'A')
a++;
else if (s.at(i).at(0) == 'R')
r++;
else if (s.at(i).at(0) == 'C')
c++;
else if (s.at(i).at(0) == 'H')
h++;
}
unsigned long long ans;
ans = m * a * r + m * a * c + m * a * h + m * r * c + m * r * h + m * c * h +
a * r * c + a * r * h + a * c * h + r * c * h;
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
k = 0;
cin >> n;
do {
string x, c;
cin >> x;
c = x[0];
if (c == "M" || c == "A" || c == "R" || c == "C" || c == "H") {
k = k + 1;
};
n = n - 1;
} while (n >= 1);
if (k <= 2) {
cout << 0;
} else {
unsigned long long m;
m = k * (k - 1) * (k - 2) / 6;
cout << m;
};
system("pause");
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, su = 0;
map<char, int> m;
string s;
cin >> n;
getchar();
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M' || s[0] == 'A' || s[0] == 'R' || s[0] == 'C' ||
s[0] == 'H') {
m[s[0]]++;
}
}
if (m['M'] != 0 && m['A'] != 0 && m['R'] != 0) {
su += max({m['M'], m['A'], m['R']});
}
if (m['M'] != 0 && m['C'] && m['H']) {
su += max({m['M'], m['C'], m['H']});
}
if (m['M'] != 0 && m['A'] && m['H']) {
su += max({m['M'], m['A'], m['H']});
}
if (m['M'] != 0 && m['R'] && m['H']) {
su += max({m['M'], m['H'], m['R']});
}
if (m['M'] != 0 && m['R'] && m['C']) {
su += max({m['M'], m['R'], m['C']});
}
if (m['M'] != 0 && m['C'] && m['A']) {
su += max({m['M'], m['C'], m['A']});
}
if (m['R'] != 0 && m['A'] != 0 && m['C'] != 0) {
su += max({m['A'], m['R'], m['C']});
}
if (m['A'] != 0 && m['R'] && m['H']) {
su += max({m['A'], m['R'], m['H']});
}
if (m['C'] != 0 && m['A'] != 0 && m['H'] != 0) {
su += max({m['A'], m['C'], m['H']});
}
if (m['H'] != 0 && m['C'] != 0 && m['R'] != 0) {
su += max({m['R'], m['C'], m['H']});
}
{ cout << su << endl; }
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, c[5] = {0};
cin >> n;
string s = "MARCH";
for (int i = 0; i < n; i++) {
string t;
cin >> t;
for (int j = 0; j < 5; j++) {
if (t[0] == s[i]) {
c[j]++;
}
}
}
int ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += c[i] * c[j] * c[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n;
int a[5];
string s;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cin >> n;
for (int i = (0); i < (n); ++i) {
cin >> s;
switch (s[0]) {
case 'M':
a[0]++;
break;
case 'A':
a[1]++;
break;
case 'R':
a[2]++;
break;
case 'C':
a[3]++;
break;
case 'H':
a[4]++;
break;
}
}
long long ans = 0;
for (int i = (0); i < (5); ++i)
for (int j = (i + 1); j < (5); ++j)
for (int k = (j + 1); k < (5); ++k) ans += a[i] * a[j] * a[k];
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
unsigned long long N, M, A, R, C, H;
int main(void) {
cin >> N;
for (int i = 0; i < N; i++) {
string str;
cin >> str;
switch (str[0]) {
case 'M':
M++;
break;
case 'A':
A++;
break;
case 'R':
R++;
break;
case 'C':
C++;
break;
case 'H':
H++;
break;
}
}
unsigned long long L[5];
L[0] = M;
L[1] = A;
L[2] = R;
L[3] = C;
L[4] = H;
int ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += L[i] * L[j] * L[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using ll = long long;
using namespace std;
int main() {
int N;
string S;
cin >> N;
map<char, int> mp;
string ch = "MARCH";
for (ll i = 0; i < N; i++) {
cin >> S;
mp[S[0]]++;
}
ll ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += mp[ch[i]] * mp[ch[j]] * mp[ch[k]];
}
}
}
cout << ans << "\n";
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace Atcoder
{
class Program
{
static void Main(string[] args)
{
//var line1 = Array.ConvertAll(Console.ReadLine().Split(), long.Parse);
//var N = line1[0];
//var M = line1[1];
//var a = new long[M];
//var b = new long[M];
//int bridge = 0;
//int[,] dag = new int[N,N];
//for (int i = 0; i < M; i++)
//{
// var line2 = Array.ConvertAll(Console.ReadLine().Split(), long.Parse);
// a[i] = line2[0]-1;
// b[i] = line2[1]-1;
// dag[a[i], b[i]] = 1;
// dag[b[i], a[i]] = 1;
//}
//for (int i = 0; i < M; i++)
//{
// bool[] visited = new bool[N];
// dag[a[i], b[i]] = 0;
// dag[b[i], a[i]] = 0;
// DFS(0,visited,dag);
// if (visited.Where(n => n == true).Count() != N)
// {
// bridge++;
// }
// dag[a[i], b[i]] = 1;
// dag[b[i], a[i]] = 1;
//}
//Console.WriteLine(bridge);
//var line1 = Array.ConvertAll(Console.ReadLine().Split(), long.Parse);
//var A = line1[0];
//var B = line1[1];
//var C = line1[2];
//var D = line1[3];
//var E = line1[4];
//var F = line1[5];
var N = int.Parse(Console.ReadLine());
char[] mozi = new char[N];
int[] MARCH = new int[5];
int cnt = 0;
for (int i = 0; i < N; i++)
{
mozi[i] = Console.ReadLine().ToCharArray()[0];
}
MARCH[0] = mozi.Where(n => n == 'M').Count();
MARCH[1] = mozi.Where(n => n == 'A').Count();
MARCH[2] = mozi.Where(n => n == 'R').Count();
MARCH[3] = mozi.Where(n => n == 'C').Count();
MARCH[4] = mozi.Where(n => n == 'H').Count();
for (int i = 0; i < 3; i++)
{
for (int j = i + 1; j < 4; j++)
{
for (int k = j + 1; k < 5; k++)
{
cnt += MARCH[i] * MARCH[j] * MARCH[k];
}
}
}
Console.WriteLine(cnt);
}
public static void DFS(int v,bool[] visited,int[,] dag)
{
visited[v] = true;
for (int i = 0; i < visited.Length; i++)
{
if (dag[v, i] == 0) continue;
if (visited[i]) continue;
DFS(i,visited,dag);
}
}
public static long LowerBound<T>(T[] arr, long start, long end, T value, IComparer<T> comparer)
{
long low = start;
long high = end;
long mid;
while (low < high)
{
mid = ((high - low) >> 1) + low;
if (comparer.Compare(arr[mid], value) < 0)
low = mid + 1;
else
high = mid;
}
return low;
}
//引数省略のオーバーロード
public static long LowerBound<T>(T[] arr, T value) where T : IComparable
{
return LowerBound(arr, 0, arr.Length, value, Comparer<T>.Default);
}
public static long UpperBound<T>(T[] arr, long start, long end, T value, IComparer<T> comparer)
{
long low = start;
long high = end;
long mid;
while (low < high)
{
mid = ((high - low) >> 1) + low;
if (comparer.Compare(arr[mid], value) <= 0)
low = mid + 1;
else
high = mid;
}
return low;
}
//引数省略のオーバーロード
public static long UpperBound<T>(T[] arr, T value)
{
return UpperBound(arr, 0, arr.Length, value, Comparer<T>.Default);
}
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | package main
import (
"bufio"
"fmt"
"os"
)
func main() {
var N int
fmt.Scan(&N)
sc := bufio.NewScanner(os.Stdin)
sc.Split(bufio.ScanWords)
h := make([][]string, N)
var a [][]string
for range h {
sc.Scan()
w := sc.Text()
switch w[:1] {
case "M":
h[0] = append(h[0], w)
case "A":
h[1] = append(h[1], w)
case "R":
h[2] = append(h[2], w)
case "C":
h[3] = append(h[3], w)
case "H":
h[4] = append(h[4], w)
}
}
for i := 0; i < len(h); i++ {
if len(h[i]) > 0 {
a = append(a, h[i])
}
}
if len(a) < 2 {
fmt.Println(0)
return
}
if len(a) == 3 {
r := 1
for i := 0; i < len(a); i++ {
r *= len(a[i])
}
fmt.Println(r)
return
}
if len(a) == 4 {
r := 4
for i := 0; i < len(a); i++ {
if len(a[i]) > 1 {
r += 3 * (len(a[i]) - 1)
}
}
fmt.Println(r)
return
}
if len(a) == 5 {
r := 10
for i := 0; i < len(a); i++ {
if len(a[i]) > 1 {
r += 6 * (len(a[i]) - 1)
}
}
fmt.Println(r)
return
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
map<char, int> head;
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'M')
head['m'] += 1;
else if (s[0] == 'A')
head['a'] += 1;
else if (s[0] == 'R')
head['r'] += 1;
else if (s[0] == 'C')
head['c'] += 1;
else if (s[0] == 'H')
head['h'] += 1;
}
char idx[5] = {'m', 'a', 'r', 'c', 'h'};
int sele1[] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
int sele2[] = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
int sele3[] = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
int sum = 0;
for (int i = 0; i < 10; i++) {
sum += head[idx[sele1[i]]] * head[idx[sele2[i]]] * head[idx[sele3[i]]];
}
cout << sum << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | import itertools
import collections
def main():
n = int(input())
S = [input()[0] for _ in range(n)]
array = ['M','A','R','C','H']
S_after = []
for s in S:
if s in array:
S_after.append(s)
if len(S_after) < 3:
print(0)
combs = itertools.combinations(set(S_after), 3)
coll = collections.Counter(S_after)
ans = 0
for comb in combs:
if comb[0] != comb[1] != comb[2]:
ans += coll[comb[0]]*coll[comb[1]]*coll[comb[2]]
print(ans)
if __name__ == '__main__':
main()
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
long[] MARCH = new long[5];
for(int i = 0; i < N; i++) {
String S = sc.next();
if(S.charAt(0) == 'M') {
MARCH[0]++;
} else if(S.charAt(0) == 'A') {
MARCH[1]++;
} else if(S.charAt(0) == 'R') {
MARCH[2]++;
} else if(S.charAt(0) == 'C') {
MARCH[3]++;
} else if(S.charAt(0) == 'H') {
MARCH[4]++;
}
}
long total = 0;
for(int i = 0; i < N - 2; i++) {
for(int j = i + 1; j < N - 1; j++) {
for(int k = j + 1; k < N; k++) {
total += MARCH[i] * MARCH[j] * MARCH[k];
}
}
}
System.out.println(total);
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
N = int(input())
S = []
for i in range(N):
S.append(input())
m = sum([name[0] == 'M' for name in S])
a = sum([name[0] == 'A' for name in S])
r = sum([name[0] == 'R' for name in S])
c = sum([name[0] == 'C' for name in S])
h = sum([name[0] == 'H' for name in S])
print(m * (a * (r + c + h) + r * (c + h) + c * h) + a * (r * c + c * h) + r * c * h)
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<string> s;
char h[5] = {'A', 'C', 'H', 'M', 'R'};
long long a, b, c, d, e, ans = 0;
void f(int i, int m) {
for (int j = i + 1; j < a; j++) {
for (int q = m + 1; q < 5; q++) {
if (s[j][0] == h[q]) {
for (int z = j + 1; z < a; z++) {
for (int x = q + 1; x < 5; x++) {
if (s[z][0] == h[x]) {
ans++;
}
}
}
}
}
}
}
int main() {
cin >> a;
string q;
for (int i = 0; i < a; i++) {
cin >> q;
q = q[0];
s.push_back(q);
}
sort(s.begin(), s.end());
for (int i = 0; i < a - 2; i++) {
for (int j = 0; j < 5; j++) {
if (s[i][0] == h[j]) {
f(i, j);
j = 5;
}
}
}
cout << ans;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int j;
int i, s1, s2, s3, s4, s5;
s1 = s2 = s3 = s4 = s5 = 0;
cin >> j;
string a;
for (i = 0; i < j; i++) {
cin >> a;
if (a[0] == 'M') s1++;
if (a[0] == 'A') s2++;
if (a[0] == 'R') s3++;
if (a[0] == 'C') s4++;
if (a[0] == 'H') s5++;
}
cout << s1 * s2 * s3 + s1 * s2 * s4 + s1 * s2 * s5 + s1 * s3 * s4 +
s1 * s3 * s5 + s1 * s4 * s5 + s2 * s3 * s4 + s2 * s3 * s5 +
s2 * s4 * s5 + s3 * s4 * s5
<< endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
set<char> cset = {'M', 'A', 'R', 'C', 'H'};
map<char, long int> m;
for (int i = 0; i < N; ++i) {
string s;
cin >> s;
if (cset.find(s[0]) != cset.end()) {
if (m.find(s[0]) == m.end())
m[s[0]] = 1;
else
m[s[0]]++;
}
}
vector<int> v;
for (auto x : m) v.push_back(x.second);
long int ans;
for (int i = 0; i < v.size(); ++i) {
for (int j = i + 1; j < v.size(); ++j) {
for (int k = j + 1; k < v.size(); ++k) {
ans += v[i] * v[j] * v[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] count = new int[5];
for (int i = 0; i < n; i++) {
String str = String.valueOf(sc.next().charAt(0));
if (str.equals("M")) {
count[0]++;
} else if (str.equals("A")) {
count[1]++;
} else if (str.equals("R")) {
count[2]++;
} else if (str.equals("C")) {
count[3]++;
} else if (str.equals("H")) {
count[4]++;
}
}
int ans = 0;
for (int i = 0; i < 3; i++) {
for (int j = i+1; j < 4; j++) {
for (int k = j+1; k < 5; k++) {
ans += count[i] * count[j] * count[k];
}
}
}
System.out.println(ans);
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python2 | # -*- coding: utf-8 -*-
import itertools
def filter_name(n):
return n.startswith("M") or n.startswith("A") or n.startswith("R") or n.startswith("C") or n.startswith("H")
def filter_ok(l):
dic = {'M': 0, 'A': 0, 'R': 0, 'C': 0, 'H': 0}
for i in l:
dic[i[0:1]] += 1
for num in dic.values():
if num >= 2:
return False
return True
n = input()
a = []
for i in range(n):
a.append(raw_input())
r = filter(filter_name, a)
c = list(itertools.combinations(r, 3))
r = filter(filter_ok, c)
print len(r)
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define tr(container, it) \
for (auto it = container.begin(); it != container.end(); it++)
#define scontains(c, x) ((c).find(x) != (c).end()) //O(log n)
#define contains(c, x) (find((c).begin(),(c).end(),x) != (c).end()) //O(n)
#define ill(_x) ll _x;scanf("%lld",&_x);
#define idb(_x) double _x;scanf("%lf",&_x);i
#define all(x) (x).begin(),(x).end()
#define pll pair<ll,ll>
#define mll map<ll,ll>
#define vll vector<ll>
#define sll set<ll>
#define vs vector<string>
#define in0(x, a, b)((x)>=a && (x)<=b )
#define in1(x, a, b)((x)>a && (x)<b)
#define rep(i, begin, end) for (__typeof(end) i = (begin); i != (end); i += 1 - 2 * ((begin) > (end)))
#define int int_64
#define endl '\n'
const double pi = 3.14159265358979323846;
const int INF = 0x3f3f3f3f;
const int X10 = 1024, X11 = 2048, X12 = 4096, X13 = 8196, X14 = 16392, X15 = 32786, X16 = 65536, X17 = 131072, X18 = 262144, X19 = 524288, X20 = 1048576;
const int MOD = (int) (1e9 + 7);
template<typename FI>
void parted_rotate(FI first1, FI last1, FI first2, FI last2)
{
if(first1 == last1 || first2 == last2) return;
FI next = first2;
while (first1 != next) {
std::iter_swap(first1++, next++);
if(first1 == last1) first1 = first2;
if (next == last2) {
next = first2;
} else if (first1 == first2) {
first2 = next;
}
}
}
template<typename BI>
bool next_combination_imp(BI first1, BI last1, BI first2, BI last2)
{
if(first1 == last1 || first2 == last2) return false;
auto target = last1; --target;
auto last_elem = last2; --last_elem;
// find right-most incrementable element: target
while(target != first1 && !(*target < *last_elem)) --target;
if(target == first1 && !(*target < *last_elem)) {
parted_rotate(first1, last1, first2, last2);
return false;
}
// find the next value to be incremented: *next
auto next = first2;
while(!(*target < *next)) ++next;
std::iter_swap(target++, next++);
parted_rotate(target, last1, next, last2);
return true;
}
// INVARIANT: is_sorted(first, mid) && is_sorted(mid, last)
template<typename BI>
inline bool next_combination(BI first, BI mid, BI last)
{
return next_combination_imp(first, mid, mid, last);
}
// INVARIANT: is_sorted(first, mid) && is_sorted(mid, last)
template<typename BI>
inline bool prev_combination(BI first, BI mid, BI last)
{
return next_combination_imp(mid, last, first, mid);
}
bool isPrime(int n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n%2 == 0 || n%3 == 0) return false;
for (int i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return false;
return true;
}
int A[5];
void _() {
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
string s;
cin >> s;
if(s[0] == 'M')A[0] ++;
if(s[0] == 'A')A[1] ++;
if(s[0] == 'R')A[2] ++;
if(s[0] == 'C')A[3] ++;
if(s[0] == 'H')A[4] ++;
}
ll sum = 0;
rep(i,0,5)rep(j,i+1,5)rep(k,j+1,5)sum += A[i]*A[j]*A[k];
cout << sum <<endl;
}
#undef int
int main() {
#ifdef Debug
freopen("/home/hiroo/Desktop/competitive-programming/IO/Input.txt", "r", stdin);
freopen("/home/hiroo/Desktop/competitive-programming/IO/Output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL);
_();
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
int D[5] = {0};
for (int i = 0; i < N; i++) {
string S;
cin >> S;
if (S[0] == 'M')
D[0]++;
else if (S[0] == 'A')
D[1]++;
else if (S[0] == 'R')
D[2]++;
else if (S[0] == 'C')
D[3]++;
else if (S[0] == 'H')
D[4]++;
}
long long comb = 0;
int P[10] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
int Q[10] = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
int R[10] = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
for (int i = 0; i < 10; i++) comb += D[P[i]] * D[Q[i]] * D[R[i]];
cout << comb << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n;
long long ans, a[10];
char s[101010];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%s", s);
if (s[0] == 'M') a[1] = 1;
if (s[0] == 'A') a[2] = 1;
if (s[0] == 'R') a[3] = 1;
if (s[0] == 'C') a[4] = 1;
if (s[0] == 'H') a[5] = 1;
}
for (int i = 1; i <= 3; i++)
for (int j = i + 1; j <= 4; j++)
for (int k = j + 1; k <= 5; k++) ans += a[i] * a[j] * a[k];
printf("%lld", ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, march[5] = {}, ans = 1;
bool exist = false;
string s;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> s;
switch (s[0]) {
case 'M':
march[0]++;
break;
case 'A':
march[1]++;
break;
case 'R':
march[2]++;
break;
case 'C':
march[3]++;
break;
case 'H':
march[4]++;
break;
default:
break;
}
}
for (int i = 0; i < 5; i++) {
if (march[i]) {
exist = true;
ans *= march[i];
}
}
cout << (exist ? ans : 0) << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | N = int(input())
Sdic = {}
L = ["M","A","R","C","H"]
for i in L:
Sdic[i] = []
tN = N
for i in range(N):
s = input()
if s[0] in L:
Sdic[s[0]].append(s)
else:
tN -= 1
zen = tN*(tN-1)*(tN-2) / 6
one = 0
for i in ["M","A","R","C","H"]:
c = len(Sdic[i])
if c >= 3:
one += c * (c-1) * (c-2) / 6
lis = []
for i in range(len(L)):
for j in range(len(L)):
if i < j:
lis.append(len(Sdic[L[i]]) + len(Sdic[L[j]]))
two = 0
for c in lis:
if c >= 3:
two += c * (c-1) * (c-2) / 6
print(int(zen - one - two))
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define leading zero str.erase(0, min(str.find_first_not_of('0'), str.size()-1));
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
string text="abcdefghijklmnopqrstuvwxyz";
const int maxn=1e6+7;
// .--------------.
// | Try First One|
// '--------------'
// | .--------------.
// | | |
// V V |
// .--------------. |
// | AC. |<---. |
// '--------------' | |
// (True)| |(False) | |
// .--------' | | |
// | V | |
// | .--------------. | |
// | | Try Again |----' |
// | '--------------' |
// | |
// | .--------------. |
// '->| Try Next One |-------'
// '--------------'
ll bin_pow(ll a,ll b,ll m)
{
ll res=1;
a%=m;
while(b>0)
{
if(b&1)
res=res*a%m;
b>>=1;
a=a*a%m;
}
return res;
}
bool miller_rabin(ll d,ll n)
{
ll a=2+rand()%(n-4);
ll x=bin_pow(a,d,n);
if(x==1 || x==n-1)
return true;
while(d!=n-1)
{
x=(x*x)%n;
d*=2;
if(x==1)
return false;
if(x==n-1)
return true;
}
return false;
}
bool prime(ll n,ll k)
{
if(n==1 || n==4)
return false;
if(n<=3)
return true;
ll d=n-1;
while(d%2==0)
d/=2;
for(int i=0; i<k; i++)
{
if(!miller_rabin(d,n))
return false;
}
return true;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin>>n;
int cnt[26]={0};
for(int i=0;i<n;i++){
string s;
cin>>s;
if(s[0]=='M')cnt[s[0]-'A']++;
if(s[0]=='A')cnt[s[0]-'A']++;
if(s[0]=='R')cnt[s[0]-'A']++;
if(s[0]=='C')cnt[s[0]-'A']++;
if(s[0]=='H')cnt[s[0]-'A']++;
}
vector<int>v;
for(int i=0;i<26;i++){
if(cnt[i]!=0)v.push_back(cnt[i]);
}
int um=accumulate(v.begin(),v.end(),0,[&](int k,int i){return k=k+i;});
if(v.size()==0 || um<3)return 0*puts("0");
ll sum=0;
for(int i=0;i<v.size()-2;i++){
for(int j=i+1;j<v.size()-1;j++){
for(int k=j+1;k<v.size();k++){
int x=v[i];int y=v[j];int z=v[k];
sum+=(x*y*z);
}
}
}
cout<<sum<<endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #include <bits/stdc++.h>
int main(void) {
int n, m = 0, a = 0, r = 0, c = 0, h = 0;
scanf("%d", &n);
char s[n][12];
for (int i = 0; i < n; i++) {
scanf("%s", s[i]);
if (s[i][0] == 'M') {
m++;
} else if (s[i][0] == 'A') {
a++;
} else if (s[i][0] == 'R') {
r++;
} else if (s[i][0] == 'C') {
c++;
} else if (s[i][0] == 'H') {
h++;
}
}
long long ans = 0;
ans += m * a * r;
ans += m * a * c;
ans += m * a * h;
ans += m * r * c;
ans += m * r * h;
ans += m * c * h;
ans += a * r * c;
ans += a * r * h;
ans += a * c * h;
ans += r * c * h;
printf("%lld\n", ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | N = int(input())
S = [input() for i in range(N)]
S_first = [S[i][0] for i in range(N)]
march = ['M','A','R','C','H']
march_number = [0]*5
for i in range(N):
march_number[i] = S_first.count(march[i])
import itertools
ptr = list(itertools.combinations(march_number,3))
ans = 0
seki = 1
for i in range(len(ptr)):
for j in range(3):
seki *= ptr[i][j]
ans += seki
seki = 1
print(ans) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | n = int(input())
l_ = [["M", 0], ["A", 0], ["R", 0], ["C", 0], ["H", 0]]
l = []
for i in range(n):
s = input()
for j in l_:
if s[0] == j[0]:
j[1] += 1
break
for i in l_:
if i[1] != 0:
l.append(i[1])
k = len(l)
r = 0
if k < 3:
r = 0
elif k == 3:
r = l[0] * l[1] * l[2]
elif k == 4:
for i in range(1,5):
r += l[i%4] * l[i%3] * l[i%2]
elif k == 5:
r = 1
for i in range(5):
r *= l[i]
for i in range(5):
for j in range(5):
r /= l[i] + l[j]
print(r) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | N = int(input())
s = []
for i in range(N):
s.append(input())
m = 0
a = 0
r = 0
c = 0
h = 0
for person in s:
if person[0]=='M':
m += 1
elif person[0]=='A':
a += 1
elif person[0]=='R':
r += 1
elif person[0]=='C':
c += 1
elif person[0]=='H':
h += 1
ans = m*a*r+m*a*c+m*a*h+m*r*c+m*r*h+m*c*h+a*r*c+a*r*h+a*c*h*r*c*h
print(ans) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007LL;
const long long INF = LLONG_MAX;
signed main() {
long long n;
cin >> n;
long long cnt[5] = {};
for (long long i = 0; i < n; i++) {
string s;
cin >> s;
if (s[i] == 'M') cnt[0]++;
if (s[i] == 'A') cnt[1]++;
if (s[i] == 'R') cnt[2]++;
if (s[i] == 'C') cnt[3]++;
if (s[i] == 'H') cnt[4]++;
}
long long ans = 0;
for (long long i = 0; i < 5; i++) {
for (long long j = i + 1; j < 5; j++) {
for (long long k = j + 1; k < 5; k++) {
ans += cnt[i] * cnt[j] * cnt[k];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main()
{
int s=0,n,i,a=0,b=0,c=0,d=0,e=0;
char a,tmp[100];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%c",&a);
scanf("%s",tmp);
if(a=='M')
e++;
if(a=='A')
a++
if(a=='R')
b++
if(a=='C')
c++;
if(a=='H')
d++;
}
s+=a*b*c+a*b*d+a*b*e+a*c*d+a*c*e+a*d*e+b*c*d+b*c*e+b*d*e+c*d*e;
printf("%d",s);
return 0;
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<int> distan(90001, 0);
int main(void) {
vector<int> arr(5);
int N;
cin >> N;
for (int i = 0; i < N; i++) {
string a;
cin >> a;
if (a[0] == 'M') {
arr[0]++;
}
if (a[0] == 'A') {
arr[1]++;
}
if (a[0] == 'R') {
arr[2]++;
}
if (a[0] == 'C') {
arr[3]++;
}
if (a[0] == 'H') {
arr[4]++;
}
}
int count = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int l = j + 1; l < 5; l++) {
count += arr[i] * arr[j] * arr[l];
}
}
}
cout << count << "\n";
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | N=int(input())
a=[]
for i in range(N):
a.append(input())
import numpy as np
s=np.zeros(5)
for i in a:
if i[0]=="M":
s[0]+=1
elif i[0]=="A":
s[1]+=1
elif i[0]=="R":
s[2]+=1
elif i[0]=="C":
s[3]+=1
elif i[0]=="H":
s[4]+=1
else:
continue
b=np.unique(s)
c=0
if len(b)<3:
print(0)
else:
for i in range(0,3):
for j in range(i+1,4):
for k in range(j+1,5):
c+=s[i]*s[j]*s[k]
print(int(c))
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | ni = lambda: int(input())
nm = lambda: map(int, input().split())
nl = lambda: list(map(int, input().split()))
n=ni()
m=0
a=0
r=0
c=0
h=0
for i in range(n):
k=input()[0]
if k=='M':
m+=1
if k=='A':
a+=1
if k=='R':
r+=1
if k=='C':
c+=1
if k=='H':
h+=1
cnts = [m,a,r,c,h]
ans=0
for i in range(5):
for j in range(i+1,5):
for k in range(j+1,5):
cs = [cnts[i],cnts[j],cnts[k]]
if 0 in cs:
continue
cnt = sum(cs)
tot = (cnt*(cnt-1)*(cnt-2))//6
for l in range(3):
if cs[l]>0:
tot//=cs[l]
ans += tot
print(ans)
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int q[6];
int main() {
int n, ans = 1;
char ch[11];
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%s", ch);
if (ch[0] == 'M') q[1]++;
if (ch[0] == 'A') q[2]++;
if (ch[0] == 'R') q[3]++;
if (ch[0] == 'C') q[4]++;
if (ch[0] == 'H') q[5]++;
}
for (int i = 1; i <= 5; i++) {
if (q[i] == 0)
;
else
ans = ans * q[i];
}
printf("%d", ans);
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int x;
cin >> x;
vector<int> vec(5, 0);
for (int i = 0; i < x; i++) {
string s;
cin >> s;
if (s.at(0) == 'M') {
vec.at(0)++;
} else if (s.at(0) == 'A') {
vec.at(1)++;
} else if (s.at(0) == 'R') {
vec.at(2)++;
} else if (s.at(0) == 'C') {
vec.at(3)++;
} else if (s.at(0) == 'H') {
vec.at(4)++;
}
}
long long ans = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
ans += vec.at(i) * vec.at(j) * vec.at(k);
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.Scanner;
import java.util.Arrays;
import java.util.ArrayDeque;
import java.util.Queue;
import java.util.LinkedList;
public class Main {
public static void main(String[] args) {
new Main().solve();
}
void solve() {
Scanner sc = new Scanner(System.in);
int N=sc.nextInt();
long ans=0;
String S[]=new String[N];
int s[]=new int[5];
Arrays.fill(s,0);
for(int i=0;i<N;i++){
S[i]=sc.next();
}
for(int i=0;i<N;i++){
if(S[i].charAt(0)=='M')s[0]++;
if(S[i].charAt(0)=='A')s[1]++;
if(S[i].charAt(0)=='R')s[2]++;
if(S[i].charAt(0)=='C')s[3]++;
if(S[i].charAt(0)=='H')s[4]++;
}
for(int i=0;i<5;i++){
for(int j=i+1;j<5;j++){
for(int k=j+1;k<5;k++){
ans+=s[i]*s[j]*s[k];
}
}
}
System.out.println(ans);
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e4 + 5;
int a[100];
int main() {
int n;
while (~scanf("%d", &n)) {
memset(a, 0, sizeof(a));
string s;
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M')
a[1]++;
else if (s[0] == 'A')
a[2]++;
else if (s[0] == 'R')
a[3]++;
else if (s[0] == 'C')
a[4]++;
else if (s[0] == 'H')
a[5]++;
}
int sum = 0, ans = 0, k = 0;
for (int i = 1; i < 6; i++) {
if (a[i] == 1)
sum++;
else if (a[i] != 0)
ans++;
k += a[i];
if (a[i] == 2) {
a[i * 10 + 2] = a[i] * (a[i] - 1) / 2;
} else if (a[i] >= 3)
a[i * 10 + 3] = a[i] * (a[i] - 1) * (a[i] - 2) / 6;
}
if (ans + sum < 3 || n < 3)
printf("0\n");
else {
int cnt = k * (k - 1) * (k - 2) / 6;
for (int i = 1; i < 6; i++) {
cnt = cnt - a[i * 10 + 3];
cnt -= sum * a[i * 10 + 2];
}
for (int i = 1; i < 6; i++) {
for (int j = 1; j < 6 && i != j && a[i] != 1; j++) {
cnt -= a[i] * a[j * 10 + 2];
}
}
printf("%d\n", cnt);
}
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #include <bits/stdc++.h>
int main() {
int n, ini[5] = {};
long double ans = 0;
char name[11];
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%s", name);
switch (name[0]) {
case 'M':
ini[0]++;
break;
case 'A':
ini[1]++;
break;
case 'R':
ini[2]++;
break;
case 'C':
ini[3]++;
break;
case 'H':
ini[4]++;
break;
default:
break;
}
}
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
ans += ini[i] * ini[j] * ini[k];
}
}
}
printf("%lld", (long long int)ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
int main() {
int N, count[5] = {
0,
};
scanf("%d", &N) + 1;
char str[16];
for (int i = 0; i < N; i++) {
scanf("%s", str) + 1;
switch (str[0]) {
case 'M':
count[0]++;
break;
case 'A':
count[1]++;
break;
case 'R':
count[2]++;
break;
case 'C':
count[3]++;
break;
case 'H':
count[4]++;
break;
default:
break;
}
}
int b0 = 1, b1 = 2, b2 = 4, b3 = 8, b4 = 16;
long long ans = 0;
for (int i = 0; i < 32; i++) {
int b = 0;
if (i & b0) b++;
if (i & b1) b++;
if (i & b2) b++;
if (i & b3) b++;
if (i & b4) b++;
if (b != 3) continue;
long long total = 1;
if (i & b0) total *= count[0];
if (i & b1) total *= count[1];
if (i & b2) total *= count[2];
if (i & b3) total *= count[3];
if (i & b4) total *= count[4];
ans += total;
}
printf("%d\n", ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <iostream>
#include <map>
#include <string>
using namespace std;
int main(){
map<int,int> name;
int n;
cin>>n;
for(int i=0;i<n;i++){
string s;
cin>>s;
name[s[0]]+=1;
}
int ans=0;
ans+=name["M"]*name["A"]*name["R"]+name["M"]*name["A"]*name["C"]+name["M"]*name["A"]*name["H"]+name["M"]*name["R"]*name["C"]+name["M"]*name["R"]*name["H"]+name["M"]*name["C"]*name["H"]+name["A"]*name["R"]*name["C"]+name["A"]*name["R"]*name["H"]+name["A"]*name["C"]*name["H"]+name["R"]*name["C"]*name["H"];
cout<<ans<<endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | # include < cstdio >
# include < iostream >
using namespace std ;
typedef long long ll ;
string s ;
int N ;
ll m ,a ,r ,c , h ;
ll D [5];
int P [10]={0 ,0 ,0 ,0 ,0 ,0 ,1 ,1 ,1 ,2};
int Q [10]={1 ,1 ,1 ,2 ,2 ,3 ,2 ,2 ,3 ,3};
int R [10]={2 ,3 ,4 ,3 ,4 ,4 ,3 ,4 ,4 ,4};
int main ()
{
scanf ("% d " ,& N );
for ( int i =0; i < N ; i ++)
{
cin > > s ;
if ( s [0]== ’ M ’) m ++;
if ( s [0]== ’ A ’) a ++;
if ( s [0]== ’ R ’) r ++;
if ( s [0]== ’ C ’) c ++;
if ( s [0]== ’ H ’) h ++;
}
D [0]= m , D [1]= a , D [2]= r , D [3]= c , D [4]= h ;
ll res =0;
for ( int d =0; d <10; d ++)
res += D [ P [ d ]]* D [ Q [ d ]]* D [ R [ d ]];
printf ("% lld \ n " , res )
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> v(5, 0);
int N;
cin >> N;
for (int i = 0; i < N; i++) {
string st;
cin >> st;
if (st[0] == 'M') v[0] += 1;
if (st[0] == 'A') v[1] += 1;
if (st[0] == 'R') v[2] += 1;
if (st[0] == 'C') v[3] += 1;
if (st[0] == 'H') v[4] += 1;
}
vector<int> I = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
vector<int> J = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
vector<int> K = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
long long rst = 0;
for (int i = 0; i < 10; i++) {
rst += v[I[i]] * v[J[i]] * v[K[i]];
}
cout << rst << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
string s[n];
for (int i = 0; i < n; i++) cin >> s[i];
if (n < 3) {
cout << 0 << endl;
return 0;
}
int num[5];
for (int i = 0; i < 5; i++) num[i] = 0;
for (int i = 0; i < n; i++) {
if (s[i][0] == 'M') num[0]++;
if (s[i][0] == 'A') num[1]++;
if (s[i][0] == 'R') num[2]++;
if (s[i][0] == 'C') num[3]++;
if (s[i][0] == 'H') num[4]++;
}
unsigned long long ans = 0;
ans += num[0] * num[1] * num[2];
ans += num[0] * num[1] * num[3];
ans += num[0] * num[1] * num[4];
ans += num[0] * num[2] * num[3];
ans += num[0] * num[2] * num[4];
ans += num[0] * num[3] * num[4];
ans += num[1] * num[2] * num[3];
ans += num[1] * num[2] * num[4];
ans += num[1] * num[3] * num[4];
ans += num[2] * num[3] * num[4];
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int n, m = 0, a = 0, r = 0, c = 0, h = 0, ans = 0;
cin >> n;
string s;
for (int i = 0; i < n; ++i) {
cin >> s;
if (s[0] == 'M')
m++;
else if (s[0] == 'A')
a++;
else if (s[0] == 'R')
r++;
else if (s[0] == 'C')
c++;
else if (s[0] == 'H')
h++;
}
ans += m * a * r + m * a * c + m * a * h + a * r * c + a * r * h + r * c * h +
m * r * c + m * r * h + m * c * h + a * c * h;
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <iomanip>
#include <queue>
#include <deque>
#include <map>
#include <unordered_map>
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define repr(e,x) for(auto& e:x)
#define MOD 1'000'000'007
using namespace std;
typedef long long ll;
typedef long double ld;
ll N;
string S[100000];
ll cnt[5];
int main(){
cin>>N;
rep(i,N) cin>>S[i];
rep(i,N){
if(S[i][0]=='M') cnt[0]++;
if(S[i][0]=='A') cnt[1]++;
if(S[i][0]=='R') cnt[2]++;
if(S[i][0]=='C') cnt[3]++;
if(S[i][0]=='H') cnt[4]++;
}
ll ans=0;
rep(i,5-2){
for(int j=i+1;j<5-1;j++){
for(int k=j+1;k<5;k++){
ans+=cnt[i]*cnt[j]%MOD*cnt[k]%MOD;
ans%=MOD;
}
}
}
cout<<ans<<endl;
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
using Graph = vector<vector<int>>;
using M = map<int, int>;
using PQ = priority_queue<int>;
using PQG = priority_queue<int, vector<int>, greater<int>>;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const ll LINF = 1e18;
char c[] = {'M', 'A', 'R', 'C', 'H'};
int main() {
int n;
cin >> n;
string s;
map<char, int> m;
for (int i = 0; i < n; i++) {
cin >> s;
m[s[0]]++;
}
ll ans = 0;
for (int bit = 0; bit < (1 << 5); bit++) {
if (__builtin_popcount(bit) != 3) continue;
int sum = 1;
for (int i = 0; i < 5; i++)
if (bit >> i & 1) sum *= m[c[i]];
ans += sum;
}
cout << (ans) << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #[allow(unused_imports)]
use proconio::marker::{Bytes, Chars};
use proconio::{fastout, input};
#[fastout]
fn main() {
input! {
n: usize,
s: [Chars; n]
}
let (mut m, mut a, mut r, mut c, mut h) = (0, 0, 0, 0, 0);
for i in 0..n {
if s[i][0] == 'M' {
m += 1
}
if s[i][0] == 'A' {
a += 1
}
if s[i][0] == 'R' {
r += 1;
}
if s[i][0] == 'C' {
c += 1;
}
if s[i][0] == 'H' {
h += 1;
}
}
let d = [m, a, r, c, h];
let p = [0, 0, 0, 0, 0, 0, 1, 1, 1, 2];
let q = [1, 1, 1, 2, 2, 3, 2, 2, 3, 3];
let t = [2, 3, 4, 3, 4, 4, 3, 4, 4, 4];
let mut ans = 0;
for i in 0..10 {
ans += d[p[i]] * d[q[i]] * d[t[i]];
}
println!("{}", ans);
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
string S[N];
for (int i = 0; i < N; i++) cin >> S[N];
long long int count[5];
for (int i = 0; i < N; i++) {
if (S[i][0] == 'M') {
count[0] += 1;
} else if (S[i][0] == 'A') {
count[1] += 1;
} else if (S[i][0] == 'R') {
count[2] += 1;
} else if (S[i][0] == 'C') {
count[3] += 1;
} else if (S[i][0] == 'H') {
count[4] += 1;
}
}
long long int sum = 0;
for (int i = 0; i < 3; i++) {
for (int j = i + 1; j < 4; j++) {
for (int k = j + 1; k < 5; k++) {
sum += count[i] * count[j] * count[k];
}
}
}
cout << sum << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC diagnostic ignored "-Wunused-result"
using namespace std;
template <class T>
using V = vector<T>;
template <class T>
using VV = vector<vector<T>>;
template <class T>
void _R(T &x) {
cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%" PRId64, &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
_R(head);
R(tail...);
}
template <class T>
void _W(const T &x) {
cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%" PRId64, x); }
void _W(const double &x) { printf("%.16f\n", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T>
void _W(const vector<T> &x) {
for (auto i = x.begin(); i != x.end(); _W(*i++))
if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
_W(head);
putchar(sizeof...(tail) ? ' ' : '\n');
W(tail...);
}
template <class T>
inline bool chmax(T &a, const T &b) {
return b > a ? a = b, true : false;
}
template <class T>
inline bool chmin(T &a, const T &b) {
return b < a ? a = b, true : false;
}
template <class T, class F = less<T>>
void sort_uniq(vector<T> &v, F f = F()) {
sort(begin(v), end(v), f);
v.resize(unique(begin(v), end(v)) - begin(v));
}
template <class T>
using MaxHeap = priority_queue<T>;
template <class T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
const int MAX_N = 1e5 + 10;
int n;
int64_t a[5];
int main() {
R(n);
for (int i = 0; i < int(n); i++) {
string s;
R(s);
if (s[0] == 'M') a[0]++;
if (s[0] == 'A') a[1]++;
if (s[0] == 'R') a[2]++;
if (s[0] == 'C') a[3]++;
if (s[0] == 'H') a[4]++;
}
int64_t ans = 0;
for (int i = 0; i < int(5); i++) {
for (int j = (i + 1); j <= int(5); j++) {
for (int k = (j + 1); k <= int(5); k++) {
ans += a[i] * a[j] * a[k];
}
}
}
W(ans);
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
int main(void) {
int N;
std::cin >> N;
std::vector<std::string> S;
for (int i = 0; i < N; i++) {
std::string a;
std::cin >> a;
S.push_back(a);
}
int m = 0, a = 0, r = 0, c = 0, h = 0;
for (int i = 0; i < N; i++) {
if (S[i][0] == 'M') {
m = 1;
} else if (S[i][0] == 'A') {
a = 1;
} else if (S[i][0] == 'R') {
r = 1;
} else if (S[i][0] == 'C') {
c = 1;
} else if (S[i][0] == 'H') {
h = 1;
}
}
int x;
x = m + a + r + c + h;
if (x >= 3) {
std::cout << (x * (x - 1) * (x - 2)) / 6 << std::endl;
}
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long combination(int n, int r) {
if (n == r)
return (1);
else if (r == 0)
return (1);
else if (r == 1)
return (n);
else
return (combination(n - 1, r - 1) + combination(n - 1, r));
}
int main() {
long long n;
cin >> n;
char s[11];
long long a[5] = {};
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'M') {
a[0] += 1;
} else if (s[0] == 'A') {
a[1] += 1;
} else if (s[0] == 'R') {
a[2] += 1;
} else if (s[0] == 'C') {
a[3] += 1;
} else if (s[0] == 'H') {
a[4] += 1;
}
}
long long cnt = 0;
long long cnt1 = 0;
for (int i = 0; i < 5; i++) {
if (a[i] >= 1) {
cnt++;
}
if (a[i] >= 2) {
cnt1 += (a[i] - 1);
}
}
if (cnt == 3) {
cout << combination(cnt, 3) + cnt1 << endl;
} else if (cnt == 4) {
cout << combination(cnt, 3) + (cnt1 * 3) << endl;
} else if (cnt == 5) {
cout << combination(cnt, 3) + (cnt1 * 6) << endl;
} else {
cout << 0 << endl;
}
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int N;
int num[5] = {};
int main() {
cin >> N;
string name;
for (int i = 0; i < N; i++) {
cin >> name;
switch (name[0]) {
case 'M':
num[0]++;
break;
case 'A':
num[1]++;
break;
case 'R':
num[2]++;
break;
case 'C':
num[3]++;
break;
case 'H':
num[4]++;
break;
}
}
int ans = 0;
for (int i = 0; i < 5; i++) {
for (int j = i + 1; j < 5; j++) {
for (int k = j + 1; k < 5; k++) {
ans += num[i] * num[j] * num[k];
}
}
}
cout << ans << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | N = int(input())
s = [0] * N
d = {"M": 0, "A": 0, "R": 0, "C": 0, "H": 0}
for i in range(N):
s[i] = input()
if s[i][0] == "M":
d['M'] += 1
elif s[i][0] == "A":
d['A'] += 1
elif s[i][0] == "R":
d['R'] += 1
elif s[i][0] == "C":
d['C'] += 1
elif s[i][0] == "H":
d['H'] += 1
total = 1
count = 0
del_keys = []
for k, v in d.items():
if v == 0:
del_keys.append(k)
else:
total *= v
count += 1
for key in del_keys:
del d[key]
if count < 3:
print(0)
exit()
elif count == 3:
print(total)
exit()
elif count == 4:
ans = 0
for k, v in d.items():
ans += total / v
print(int(ans))
exit()
elif count == 5:
ans = 0
d_lst = list(d)
for i in range(4):
for j in range(i+1, 5):
ans += total / (d_lst[i] * d_lst[j])
print(int(ans))
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long MOD = 1000000007;
string S = "MARCH";
signed main() {
cin.tie(0);
ios::sync_with_stdio(false);
long long N;
cin >> N;
vector<string> A(N);
long long res = 0;
map<char, long long> mp;
for (long long i = 0; i < N; i++) {
cin >> A[i];
mp[A[i][0]]++;
}
for (long long i = 0; i < 5; i++) {
for (long long j = 0; j < i; j++) {
for (long long k = 0; k < j; k++) {
res = (res + mp[S[i]] * mp[S[j]] * mp[S[k]]) % MOD;
}
}
}
cout << res << endl;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5;
int main() {
int n;
char a[maxn + 5][15];
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
map<char, int> mp;
map<string, int> vis;
for (int i = 0; i < n; i++) {
string b = a[i];
mp[a[i][0]]++;
vis[b]++;
if (vis[b] == 2) {
mp[a[i][0]]--;
vis[b]--;
}
}
int pos1, pos2, pos3, pos4, pos5;
pos1 = mp['M'];
pos2 = mp['A'];
pos3 = mp['R'];
pos4 = mp['C'];
pos5 = mp['H'];
long long ans = pos1 * pos2 * pos3 + pos2 * pos3 * pos4 + pos1 * pos2 * pos5 +
pos1 * pos3 * pos4 + pos1 * pos2 * pos4 + pos2 * pos3 * pos5 +
pos3 * pos4 * pos5 + pos1 * pos3 * pos5 + pos1 * pos4 * pos5 +
pos2 * pos4 * pos5;
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | N = int(input())
A = [input() for _ in range(N)]
B = list()
num_M=0
num_A=0
num_R=0
num_C=0
num_H=0
for i in A:
if i[0]=='M':
num_M+=1
continue
if i[0]=='A':
num_A+=1
continue
if i[0]=='R':
num_R+=1
continue
if i[0]=='C':
num_C+=1
continue
if i[0]=='H':
num_H+=1
continue
print(num_M*num_A*num_R+num_M*num_A*num_C+num_M*num_A*num_H+num_M*num_R*num_C+num_M*num_R*num_H+num_M*num_C*num_H+num_A*num_R*num_C+num_A*num_R*num_H+num_R*num_C*num_H) |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N, M = 0, A = 0, R = 0, C = 0, H = 0, sum = 0;
long long int ans = 0;
cin >> N;
for (int i = 0; i < N; i++) {
string S;
cin >> S;
if (S[0] == 'M') {
if (M == 0) sum++;
M++;
}
if (S[0] == 'A') {
if (A == 0) sum++;
A++;
}
if (S[0] == 'R') {
if (R == 0) sum++;
R++;
}
if (S[0] == 'C') {
if (C == 0) sum++;
C++;
}
if (S[0] == 'H') {
if (H == 0) sum++;
H++;
}
}
if (sum >= 3) {
int k = 0;
if (sum == 3) {
ans = 1;
if (M >= 2) ans += (ans) * (M - 1);
if (A >= 2) ans += (ans) * (A - 1);
if (R >= 2) ans += (ans) * (R - 1);
if (C >= 2) ans += (ans) * (C - 1);
if (H >= 2) ans += (ans) * (H - 1);
}
if (sum == 4) {
ans = 4;
k = 1;
if (M >= 2) {
ans += (ans - k) * (M - 1);
k++;
}
if (A >= 2) {
ans += (ans - k) * (A - 1);
k++;
}
if (R >= 2) {
ans += (ans - k) * (R - 1);
k++;
}
if (C >= 2) {
ans += (ans - k) * (C - 1);
k++;
}
if (H >= 2) {
ans += (ans - k) * (H - 1);
k++;
}
}
if (sum == 5) {
ans = 10;
k = 4;
if (M >= 2) {
ans += (ans - k) * (M - 1);
k++;
}
if (A >= 2) {
ans += (ans - k) * (A - 1);
k++;
}
if (R >= 2) {
ans += (ans - k) * (R - 1);
k++;
}
if (C >= 2) {
ans += (ans - k) * (C - 1);
k++;
}
if (H >= 2) {
ans += (ans - k) * (H - 1);
k++;
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long modo(long long a) { return a % (int)(1e9 + 7); }
long long a[5];
int main() {
int ss = 0;
long long n;
scanf("%lld", &n);
;
for (int i = 0; i < n; ++i) {
string s;
std::cin >> s;
switch (s[0]) {
case 'M':
a[0]++;
break;
case 'A':
a[1]++;
break;
case 'R':
a[2]++;
break;
case 'C':
a[3]++;
break;
case 'H':
a[4]++;
break;
}
}
for (int i = 0; i < 5; ++i) {
for (int j = i + 1; j < 5; ++j) {
for (int k = j + 1; k < 5; ++k) {
ss += a[i] * a[j] * a[k];
}
}
}
std::cout << ss;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | java | import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int m = 0;
int a = 0;
int r = 0;
int c = 0;
int h = 0;
String S[] = new String [N];
for(int i=0;i<N;i++){
S[i] = String.valueOf(sc.next().charAt(0));
if(S[i].equals("M"))
m++;
else if(S[i].equals("A"))
a++;
else if(S[i].equals("R"))
r++;
else if(S[i].equals("C"))
c++;
else if(S[i].equals("H"))
h++;
}
// System.out.println(m + " " + a + " "+ r + " "+ c + " "+ h );
System.out.println(m*a*r+m*a*c+m*a*h+m*r*c+m*r*h+m*c*h+a*r*c+a*r*h+a*c*h+r*c*h);
}
} |
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python2 | import itertools
n=int(raw_input())
first_char=["M","A","R","C","H"]
name={}
for i in range(n):
s=raw_input()
name[s[0]]+=1
ans=0
for s in list(itertools.combinations(first_char,3)):
ans+=name[s[0]]*name[s[1]]*name[s[2]]
print ans
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int res = 0;
string s;
int m = 0;
int a = 0;
int r = 0;
int c = 0;
int h = 0;
for (int i = (int)(0); (i) < (int)(n); i++) {
cin >> s;
if (s[0] == 'M') m++;
if (s[0] == 'A') a++;
if (s[0] == 'R') r++;
if (s[0] == 'C') c++;
if (s[0] == 'H') h++;
}
int D[5] = {m, a, r, c, h};
int P[10] = {0, 0, 0, 0, 0, 0, 1, 1, 1, 2};
int Q[10] = {1, 1, 1, 2, 2, 3, 2, 2, 3, 3};
int R[10] = {2, 3, 4, 3, 4, 4, 3, 4, 4, 4};
for (int j = (int)(0); (j) < (int)(10); j++) {
res += D[P[j]] * D[Q[j]] * D[R[j]];
}
cout << res << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | python3 | from collections import defaultdict
if __name__ == "__main__":
dic = defaultdict(int)
N = int(input())
for i in range(N):
S = input()
if S[0] in "M A R C H":
dic[S[0]] += 1
res = []
if len(dic) < 3:
print (0)
exit(0)
value = 1
for key in dic:
res.append(dic[key])
value *= dic[key]
if len(dic) == 3:
print (value)
exit(0)
result = 0
for i in res:
result += value // i
print (result)
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
scanf("%d", &n);
int m = 0;
int a = 0;
int r = 0;
int c = 0;
int h = 0;
string str;
int valid = 0;
for (int i = 0; i < n; i++) {
cin >> str;
if (str[0] == 'M') {
m++;
valid++;
}
if (str[0] == 'A') {
a++;
valid++;
}
if (str[0] == 'R') {
r++;
valid++;
}
if (str[0] == 'C') {
c++;
valid++;
}
if (str[0] == 'H') {
h++;
valid++;
}
}
if (valid < 3) {
cout << 0 << endl;
} else {
unsigned long long sum = 0;
sum += m * a * r;
sum += m * a * c;
sum += m * a * h;
sum += m * r * c;
sum += m * r * h;
sum += m * c * h;
sum += a * r * c;
sum += a * r * h;
sum += a * c * h;
sum += r * c * h;
printf("%llu\n", sum);
}
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int N, counter = 0;
vector<string> data;
string temp;
cin >> N;
for (int i = 0; i < N; i++) {
cin >> temp;
if (temp[0] == 'M' || temp[0] == 'A' || temp[0] == 'R' || temp[0] == 'C' ||
temp[0] == 'H') {
data.push_back(temp);
}
}
if (data.size() < 3) {
cout << counter << endl;
return 0;
}
for (vector<string>::size_type i = 0; i < data.size() - 2; i++) {
for (vector<string>::size_type j = i + 1; j < data.size() - 1; j++) {
for (vector<string>::size_type k = j + 1; k < data.size(); k++) {
if (data[i][0] != data[j][0] && data[k][0] != data[j][0] &&
data[i][0] != data[k][0]) {
counter++;
}
}
}
}
cout << counter << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
map<char, int> mp{
{'M', 0}, {'A', 1}, {'R', 2}, {'C', 3}, {'H', 4},
};
bool isMARCH(char c) {
return (c == 'M' || c == 'A' || c == 'R' || c == 'C' || c == 'H');
}
int bit(int x) {
int cnt = 0;
while (x != 0) {
if (x & 1) cnt++;
x /= 2;
}
return cnt;
}
int main() {
int N;
cin >> N;
vector<int> cnt(5, 0);
for (int i = 0; i < N; i++) {
string S;
cin >> S;
if (isMARCH(S[0])) cnt[mp[S[0]]]++;
}
long long ans = 0;
for (int i = 0; i < 1 << 5; i++) {
if (bit(i) == 3) {
int x = 1;
for (int j = 0; j < 5; j++)
if ((i & 1 << j)) x *= cnt[j];
ans += x;
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(void) {
cin.tie(0);
ios::sync_with_stdio(false);
int N;
cin >> N;
map<char, long long int> ma;
string S;
for (int i = 0; i < N; i++) {
cin >> S;
ma[S[0]]++;
}
char s[5] = {'M', 'A', 'R', 'C', 'H'};
long long int ans = 0;
for (int i = 0; i <= N - 1; i++) {
for (int j = i + 1; j <= N - 1; j++) {
for (int k = j + 1; k <= N - 1; k++) {
ans += ma[s[i]] * ma[s[j]] * ma[s[k]];
}
}
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
char a[100010][100];
int n;
int an, b, c, d, e;
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
if (a[i][0] == 'M') an++;
if (a[i][0] == 'A') b++;
if (a[i][0] == 'R') c++;
if (a[i][0] == 'C') d++;
if (a[i][0] == 'H') e++;
}
int a1, b1, c1;
a1 = an * b * c + an * b * d + an * b * e + an * c * d + an * c * e +
an * d * e;
b1 = b * c * d + b * c * e + b * d * e;
c1 = c * d * e;
cout << a1 + b1 + c1 << endl;
system("pause");
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
int N;
cin >> N;
map<char, int> mp;
for (int i = 0; i < N; ++i) {
string S;
cin >> S;
mp[S[0]] = (mp.count(S[0]) != 0) ? mp[S[0]] + 1 : 1;
}
vector<int> vec(5);
vec[0] = (mp.count('M') != 0) ? mp['M'] : 0;
vec[1] = (mp.count('A') != 0) ? mp['A'] : 0;
vec[2] = (mp.count('R') != 0) ? mp['R'] : 0;
vec[3] = (mp.count('C') != 0) ? mp['C'] : 0;
vec[4] = (mp.count('H') != 0) ? mp['H'] : 0;
ll ans = 0;
for (int i = 0; i < N - 2; ++i)
for (int j = i + 1; j < N - 1; ++j)
for (int k = j + 1; k < N; ++k) {
ans += vec[i] * vec[j] * vec[k];
}
cout << ans << endl;
return 0;
}
|
p03425 AtCoder Beginner Contest 089 - March | There are N people. The name of the i-th person is S_i.
We would like to choose three people so that the following conditions are met:
* The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`.
* There are no multiple people whose names begin with the same letter.
How many such ways are there to choose three people, disregarding order?
Note that the answer may not fit into a 32-bit integer type.
Constraints
* 1 \leq N \leq 10^5
* S_i consists of uppercase English letters.
* 1 \leq |S_i| \leq 10
* S_i \neq S_j (i \neq j)
Input
Input is given from Standard Input in the following format:
N
S_1
:
S_N
Output
If there are x ways to choose three people so that the given conditions are met, print x.
Examples
Input
5
MASHIKE
RUMOI
OBIRA
HABORO
HOROKANAI
Output
2
Input
4
ZZ
ZZZ
Z
ZZZZZZZZZZ
Output
0
Input
5
CHOKUDAI
RNG
MAKOTO
AOKI
RINGO
Output
7 | {
"input": [
"5\nMASHIKE\nRUMOI\nOBIRA\nHABORO\nHOROKANAI",
"5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO",
"4\nZZ\nZZZ\nZ\nZZZZZZZZZZ"
],
"output": [
"2",
"7",
"0"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
for (auto &v : vec) is >> v;
return is;
}
template <typename A, typename B>
inline bool chmin(A &a, const B &b) {
if (a > b) {
a = b;
return true;
} else {
return false;
}
}
template <typename A, typename B>
inline bool chmax(A &a, const B &b) {
if (a < b) {
a = b;
return true;
} else {
return false;
}
}
int main() {
int N;
cin >> N;
map<char, ll> cnt;
vector<char> comb{'M', 'A', 'R', 'C', 'H'};
for (int i = 0; i < (int)(N); i++) {
string s;
cin >> s;
cnt[s[0]]++;
}
ll ans = 0;
for (int bit = 0; bit < (1 << 5); bit++) {
vector<int> idx;
for (int i = 0; i < N; i++) {
if (bit & (1 << i)) idx.push_back(i);
}
if (idx.size() != 3) continue;
ans += 1LL * cnt[comb[idx[0]]] * cnt[comb[idx[1]]] * cnt[comb[idx[2]]];
}
cout << ans << endl;
}
|
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