2024 江西省赛 ACGHJK
A. Maliang Learning Painting
思路:签到，加起来就行了。
// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
int main()
{
ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
ll a,b,c; cin>>a>>b>>c;
cout<<a+b+c<<"\n";
return 0;
}
C. Liar
思路：先看当前的和sum和正确的和s，如果是一样的那么最多所有人都是真话。如果sum>s，那么大了，要变小，由于\(-10^4\le a_i\le10^4\)。为了变的次数尽量的少，我们肯定把大的变小会更优。否则把小的变大，直到第一个能跨变的地方(原来sum<s，变到sum>=s)。
// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
ll a[N];
int main()
{
ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
ll n,s; cin>>n>>s;
ll sum = 0;
for(int i = 1;i <= n; i++)
{
ll x; cin>>x;
a[i] = x;
sum += x;
}
if(sum == s)cout<<n<<"\n";
else{
ll cnt = 0;
sort(a+1,a+1+n);
if(sum > s)
{
for(int i = n;i >= 1; i--)
{
sum -= a[i];
sum += 1e4;
cnt++;
if(sum >= s)
break;
}
}else{
for(int i = 1;i <= n; i++)
{
sum -= a[i];
sum += (-1e4);
cnt++;
if(sum <= s)
break;
}
}
cout<<n-cnt<<"\n";
}
return 0;
}
G. Multiples of 5
思路：因为是11进制，很神奇的就是，它的任意次幂的最后一位都是1，我们只需要看有多少个1然后判断就行了。
// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
int main()
{
ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
int t; cin>>t;
while(t--)
{
int n; cin>>n;
ll ans = 0;
for(int i = 1;i <= n; i++)
{
ll a; char b; cin>>a>>b;
ll t;
if(b != 'A')t = b-'0';
else t = 10;
// cout<<"t = "<<t<<"\n";
ans += a*t;
}
// cout<<ans<<"\n";
cout<<(ans%5==0?"Yes\n":"No\n");
}
return 0;
}
H.Convolution
// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 1e3 + 10;
ll I[N][N];
ll s[N][N];
int main()
{
ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
int n,m,k,l; cin>>n>>m>>k>>l;
for(int i = 1;i <= n; i++)
for(int j = 1;j <= m; j++)
cin>>I[i][j];
for(int i = 1;i <= n; i++)
{
for(int j = 1;j <= m; j++)
{
s[i][j] = s[i-1][j]+s[i][j-1]-s[i-1][j-1]+I[i][j];
}
}
ll ans = 0;
for(int i = 1;i <= k; i++)
{
for(int j = 1;j <= l; j++)
{
int x1 = i,y1 = j;
int x2 = x1 + n - k,y2 = y1 + m - l;
ll t = s[x2][y2] - s[x1-1][y2] - s[x2][y1-1] + s[x1-1][y1-1];
// cout<<t<<" ";
if(t >= 0)ans += t;
else ans -= t;
}
// cout<<"\n";
}
cout<<ans<<"\n";
return 0;
}
J.Magic Mahjong
思路：考的读题，模拟即可。
// AC one more times
// nndbk
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 10;
int cntp[N],cnts[N],cntm[N],cntz[N];
int main()
{
ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
int t; cin>>t;
while(t--)
{
memset(cntp,0,sizeof(cntp));
memset(cnts,0,sizeof(cnts));
memset(cntm,0,sizeof(cntm));
memset(cntz,0,sizeof(cntz));
string str; cin>>str;
str = "?" + str;
int idx = 1;
vector<int>p,s,m,z;
while(idx <= 28)
{
char op = str[idx+1];
// cout<<"op = "<<op<<"\n";
if(op == 'p')
p.push_back(str[idx]-'0');
else if(op == 's')
s.push_back(str[idx]-'0');
else if(op == 'm')
m.push_back(str[idx]-'0');
else if(op == 'z')
z.push_back(str[idx]-'0');
idx += 2;
}
for(auto x : p)
cntp[x]++;
for(auto x : s)
cnts[x]++;
for(auto x : m)
cntm[x]++;
for(auto x : z)
cntz[x]++;
// //p
// for(int i = 1;i <= 9; i++)
// cout<<cntp[i]<<" ";
// cout<<"\n";
// //s
// for(int i = 1;i <= 9; i++)
// cout<<cnts[i]<<" ";
// cout<<"\n";
// //m
// for(int i = 1;i <= 9; i++)
// cout<<cntm[i]<<" ";
// cout<<"\n";
// //z
// for(int i = 1;i <= 9; i++)
// cout<<cntz[i]<<" ";
// cout<<"\n";
//Thirteen Orphans
bool ok = true,other = false;
for(int i = 1 ; i <= 7; i++)
{
if(cntz[i] == 0){
ok = false;
break;
}
if(cntz[i] >= 2)other = true;
}
if(cntp[1] == 0 || cntp[9] == 0 || cnts[1] == 0 || cnts[9] == 0 || cntm[1] == 0 || cntm[9] == 0)
ok = false;
if(cntp[1] >= 2 || cntp[9] >= 2 || cnts[1] >= 2 || cnts[9] >= 2 || cntm[1] >= 2 || cntm[9] >= 2)
other = true;
if(ok && other)
{
cout<<"Thirteen Orphans\n";
continue;
}
//7 Pairs
int pair = 0;
//p
for(int i = 1;i <= 9; i++)
pair += (cntp[i] >= 2);
//s
for(int i = 1;i <= 9; i++)
pair += (cnts[i] >= 2);
//m
for(int i = 1;i <= 9; i++)
pair += (cntm[i] >= 2);
//z
for(int i = 1;i <= 9; i++)
pair += (cntz[i] >= 2);
if(pair >= 7)
cout<<"7 Pairs\n";
else cout<<"Otherwise\n";
}
return 0;
}
K. Magic Tree
其实每次到一个路口都有2中选择，答案就是\(2^{n-1}\)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 998244353;
int main(){
int m;cin >> m;
ll ant = 1;
for(int i = 0;i < m - 1;i++){
ant = (ant * 2) % MOD;
}
cout << ant;
return 0;
}