CUST新生爽爽爽部分一血代码

发表于 分类于 阅读次数： 评论数：
本文字数： 1k 阅读时长 ≈ 3 分钟
本文记录了cust新生爽爽爽的部分题解

恭喜Winners

liangxinzhu & techtbw

一血代码：

A:

Description

Solution

简单模拟。

Code

by e792a8

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
#include <iostream>
#include <cstring>
using namespace std;

int main(){
	char str[53];
	while(cin.getline(str,52)){
		for(int i=0; i<strlen(str); ++i){
			if(isupper(str[i]))
				str[i]=tolower(str[i]);
		}
		cout << str << endl;
	}
	return 0;
}

B:

Description

Solution

模拟题意。

Code

by liangxinzhu

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
typedef long long ll;
ll a, b;
int main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	while(cin >> a >> b) {
		if(a == b) {
			cout << 0 << endl;
			continue;
		}
		ll x = a - b;
		if(x < 0) cout << '-', x = -x;
		vector<char> v; int cnt = 0;
		while(x) {
			v.push_back((x % 10) + '0'), x /= 10, cnt++;
			if(cnt % 3 == 0 && x) v.push_back(',');
		}
		reverse(v.begin(), v.end());
		for(int i = 0; i < (int)v.size(); i++)
			cout << v[i];
		cout << endl;
	}
	return 0;
}

C:

Description

Solution

模拟

Code

by CUST_190331203

1
2
3
4
5
6
7
8
9
10
11
12
13
14
#include<cstdio>
#define ll long long
using namespace std;
int main(){
	char c[100];
	while(fgets(c,100,stdin)!=NULL){
		for(int i=0;c[i];++i){
			if(c[i]>='D'&&c[i]<='Z'||c[i]>='d'&&c[i]<='z')c[i]-=3;
			else if(c[i]>='A'&&c[i]<='C'||c[i]>='a'&&c[i]<='c')c[i]+=23;
			putchar(c[i]);
		}
	}	
	return 0;
}

D:

Description

Solution

Code

by CUST_190331203

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
char r[100011];
int main(){
	int a,n,e;
	while(~scanf("%d",&n)){
		memset(r,0,100010);
		for(int i=0;i<n-1;++i){
			scanf("%d",&a);
			r[a]=1;
		}
		for(int i=1;i<=n;++i){
			if(!r[i]){
				printf("%d",i);
				break;
			}
		}
		putchar('\n');
	}
	return 0;
}

E:

Description

Solution

map的简单运用

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
int main(){
	int n;
	while(~scanf("%d", &n)){
		map<string, int> mp;
		for(int i = 1; i <= n; i++) {
			string a;
			cin >> a;
			for(int j = 0; j < a.length(); j++) {
				if(a[j] >= 'A' && a[j] <= 'Z') a[j] += 32;
			}
			mp[a]++;
		}
		printf("%d\n", mp.size());
	}
	return 0;
}

F:

Description

Solution

Code

by CUST_190331203

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
int main(){
	char c[120],ps[100];
	while(fgets(ps,100,stdin)!=NULL){
		fgets(c,120,stdin);
		for(int i=0;c[i];++i){
			if(c[i]>='A'&&c[i]<='Z')c[i]=ps[c[i]-'A'];
			else if(c[i]>='a'&&c[i]<='z')c[i]=('a'-'A')+ps[c[i]-'a'];
			putchar(c[i]);
		}
		//putchar('\n');
	}	
	return 0;
}

G:

Description

Solution

Code

by 200511535_ZBW

1
2
3
4
5
6
7
8
9
10
11
12
13
14
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
int main()
{
    int x,y,k;
    while(cin>>x>>y>>k)
    {
        if(x*y*k>=100000) cout<<"1"<<'\n';
        else cout<<"0"<<'\n';
    }
    return 0;
}

H:

Description

Solution

map的简单运用

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
int main(){
	int n;
	while(~scanf("%d", &n)){
		map<int, int> mp;
		for(int i = 1; i <= n; i++) {
			int x; scanf("%d", &x);
			mp[x]++;
		}
		int ans = 0;
		map<int,int> ::iterator it;
		for(it = mp.begin(); it != mp.end(); it++) {
			ans = max(ans, it->second);
		}
		for(it = mp.begin(); it != mp.end(); it++) {
			if(it->second == ans) {
				printf("%d %d\n", it->first, it->second);
				break;
			}
		}
	}
	return 0;
}

I:

Description

Solution

Code

by techtbw

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
#include <iostream>
#include <string>
#include <vector>
using namespace std;
#define rep(i,a,b) for (int i=a;i<=b;i++)
#define per(i,a,b) for (int i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
const int N=1010;
ll x,y,z;

void solve()
{
    z=x+y;
    int res=0;
    while(z)
    {
        int t=z%10;
        z/=10;
        if(t==5) res++;
    }
    cout<<res<<endl;
}

int main()
{
    int T;cin>>T;
    while(T--)
    {
        cin>>x>>y;
        solve();
    }
	return 0;
}

J:

Description

Solution

Code

by liangxinzhu

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
string s;
ll cal(string str) {
	ll res = 0;
	for(int i = 0; i < (int)str.size(); i++) {
		res = res * 10 + str[i] - '0';
	}
	return res;
}
void doit() {
	vector<ll> v; string temp = "";
	for(int i = 0; i < (int)s.size(); i++) {
		if(s[i] == '5') {
			if(temp.size()) v.push_back(cal(temp));
			temp = "";
		}else temp += s[i];
	}
	if(temp.size()) v.push_back(cal(temp));
	sort(v.begin(), v.end());
	for(int i = 0; i < (int)v.size(); i++)
		cout << v[i] << (i != (int)v.size()-1 ? ' ' : '\n');
}
int main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	while(cin >> s) doit();
	return 0;
}

K:

Description

Solution

Code

Nobody

L:

Description

Solution

正整数唯一分解

Code

by liangxinzhu

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <cmath>
using namespace std;
typedef long long ll;
void doit() {
	int x; cin >> x;
	int sq = sqrt(x); ll ans = 1;
	for(int i = 2; i <= sq; i++) {
		int p = i, q = x / i;
		if(x % i) continue;
		ans += p;
		if(p != q) ans += q;
	}
	cout << ans << endl;
}
int main() {
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int _; cin >> _; while(_--) doit();
	return 0;
}

M:

Description

Solution

会读入就行了

Code

by CUST_190331203

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
char s[100000];
int ap[27];
int main(){
	while(fgets(s,100000,stdin)!=NULL){
		memset(ap,0,27*sizeof(int));
		for(int i=0;s[i];++i){
			if(s[i]>='a'&&s[i]<='z')
				ap[s[i]-'a']++;
		}
		for(int i=0;i<26;++i){
			printf("%c:%d\n",i+'a',ap[i]);
		}
		putchar('\n');
	}	
	return 0;
}

N:

Description

Solution

Code

Nobody

O:

Description

Solution

Code

by CUST_190331203

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
int main(){
	int a,b,c,d;
	char e[100],f[100];
	while(~scanf("%d%d%d",&a,&b,&c)){
		if(a==b&&a==0)break;
		sprintf(e,"%d",a);
		sprintf(f,"%d",b);
		for(int i=strlen(e)-1,j=strlen(f)-1;strlen(e)-i<=c;--i,--j){
			if((i>=0?e[i]:'0')!=(j>=0?f[j]:'0'))break;
			if(strlen(e)-i==c){
				printf("-1");
				goto endloop;
			}
		}
		d=a+b;
		printf("%d",d);
		endloop:
		putchar('\n');
				
	}
	return 0;
}

P:

Description

Solution

Code

by CUST_190331203

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
int main(){
	int n;
	int r[1000];
	int v;
	int ans;
	while(~scanf("%d",&n)){
		ans=0;
		if(!n)break;
		for(int i=0;i<n;++i){
			scanf("%d",r+i);
		}
		scanf("%d",&v);
		for(int i=0;i<n;++i){
			if(r[i]==v)ans++;
		}
		printf("%d\n",ans);
	}
	return 0;
}

Q:

Description

Solution

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
#include<stdio.h>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cmath>
#include<set>
#include<queue>
#define fr freopen("in.txt","r",stdin);
#define fw freopen("out.txt","w",stdout);
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
inline void print_Case(int cnt,ll ans)
{
    printf("Case #%d: %lld\n",cnt,ans);
}
int f[1000005],m,prime[50005];
set<int>s;
void primes(int n)
{
    m=0;
    for(int i=2;i<=n;i++){
        if(f[i]==0){
            f[i]=i;
            prime[++m]=i;
        }
        for(int j=1;j<=m;j++){
            if(prime[j]>f[i]||prime[j]>n/i)break;
            f[prime[j]*i]=prime[j];
        }
    }
}
const int N=5e5,M=5e6+5;
int a[5];
int main()
{
    primes(10000);
    for(int i=1;i<=m;i++)s.insert(prime[i]);
    int n;
    while(scanf("%d",&n)!=EOF){
        int a=0,b=10000;
        for(int i=1;i<=m;i++){
            if(prime[i]>n/2)break;
            if(s.find(n-prime[i])!=s.end()&&n-prime[i]-prime[i]<b-a&&n-prime[i]-prime[i]>=0){
                a=prime[i];
                b=n-prime[i];
            }
        }
        printf("%d %d\n",a,b);
    }
    return 0;
}

R:

Description

Solution

简单数学

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
void solve(int kase){
    int n;
    scanf("%d", &n);
    std::vector<int> v;
    while(1){
        if(n & 1) v.push_back(n);
        if(n & 1) n = n * 3 + 1;
        else n /= 2;
        if(n == 1) break;
    }
    if(v.size()) {
        for(int i = 0; i < v.size(); i++)
            printf("%d%c", v[i], " \n"[i == int(v.size()) - 1]);
    }else printf("No number can be output !\n");
}

S:

Description

Solution

映射关系

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
int a[3000 + 10];
int b[N];
void solve(int kase){
    int n, m;
    while(~scanf("%d%d", &n, &m)){
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        int tot = 0;
        for(int i = 1; i <= n; i++) {
            for(int j = i + 1; j <= n; j++) {
                b[++tot] = a[i] + a[j];
            }
        }
        sort(b + 1, b + tot + 1, greater<int>());
        for(int i = 1; i <= m; i++)
            printf("%d%c", b[i], " \n"[i == m]);
    }
}

T:

Description

Solution

欧拉函数

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
bool vis[N];
int pri[N], phi[N], tot;
void euler_table(int N){
    phi[1] = 1; vis[1] = vis[0] = 1;
    for(int i = 2 ;i <= N; i++){
        if(!vis[i]){
            pri[++tot] = i;
            phi[i] = i - 1;//i是素数时＜i的所有大于0的数都与i互质
        }
        for(int k = 1; k <= tot; k++){
            if(i * pri[k] > N)break;
            vis[i * pri[k]] = 1;//筛选掉i的倍数
            if(i % pri[k] == 0){//如果有一个质数是i的因子
                phi[i * pri[k]] = phi[i] * pri[k];
                break;
            }else phi[i * pri[k]] = phi[i] * phi[pri[k]];
        }
    }
}
void solve(int kase){
    int n;
    scanf("%d", &n);
    printf("%d\n", phi[n]);
}

U:

Description

Solution

模拟？

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
inline void print_Case(int cnt,ll ans)
{
    printf("Case #%d: %lld\n",cnt,ans);
}
bool judge(int n)
{
    int temp=0;
    for(int i=1;i<=n-1;i++)
    {
        if(n%i==0)temp+=i;
    }
    if(temp==n)return true;
    return false;
}
const int N=5e5,M=5e6+5;
int main()
{
    int T;
    cin>>T;
    while(T--){
        int n1,n2;
        scanf("%d%d",&n1,&n2);
        int ans=0;
        if(n1>n2)swap(n1,n2);
        if(n1<=6&&6<=n2)ans++;
        if(n1<=28&&28<=n2)ans++;
        if(n1<=496&&496<=n2)ans++;
        if(n1<=8128&&8128<=n2)ans++;
        printf("%d\n",ans);
    }
    return 0;
}

V:

Description

Solution

枚举

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
int main()
{
    int num;
    while(scanf("%d",&num)!=EOF) {
        int f=0;
        for (int i = 1; i <= 100; i++) {
            for (int j = i; j<= 100; j++) {
                for (int k = j; k <= 100; k++) {
                    if (i * i + j * j + k * k == num) {
                        printf("%d %d %d\n", i, j, k);
                        f = 1;
                        break;
                    }
                }
                if(f)break;
            }
            if(f)break;
        }
    }
    return 0;
}

W:

Description

Solution

Code

Nobody

X:

Description

Solution

进制转换

Code

by cyh20021231

1
2
3
4
5
6
7
8
9
#include <stdio.h>  
int main()  
{  
    int a, b;  
    while(scanf("%x%x", &a, &b) != EOF) {  
        printf("%d\n", a + b);  
    }  
    return 0;  
}

Y:

Description

Solution

枚举？

Code

by myfnb

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
bool judge(int n)
{
    int temp=0;
    for(int i=1;i<=n-1;i++)
    {
        if(n%i==0)temp+=i;
    }
    if(temp==n)return true;
    return false;
}
const int N=5e5,M=5e6+5;
int a[5];
int main()
{
    int f=0;
    while(1){
        vector<int>vc;
        vector<int>vc2[10];
        scanf("%d%d%d%d",&a[1],&a[2],&a[3],&a[4]);
        if(a[1]==0&&a[2]==0&&a[3]==0&&a[4]==0)break;
        if(f)printf("\n");
        else f=1;
        do{
            vc.push_back(a[1]*1000+a[2]*100+a[3]*10+a[4]);
        }while(next_permutation(a+1,a+5));
        sort(vc.begin(),vc.end());
        for(int i=0;i<vc.size();i++){
            if(vc[i]/1000==0)continue;
            vc2[vc[i]/1000].push_back(vc[i]);
        }
        for(int i=1;i<=9;i++){
            int len=vc2[i].size();
            for(int j=0;j<len;j++){
                if(j==len-1)printf("%d\n",vc2[i][j]);
                else printf("%d ",vc2[i][j]);
            }
        }
    }
    return 0;
}

Z:

Description

Solution

Code

Nobody

-------------本文结束感谢您的阅读-------------