A. Sum of Round Numbers
得到每个数位上的数是几，在记录后面零的个数
#include <bits/stdc++.h>
using namespace std;
void solve(){
int x;
cin >> x;
vector<int> a;
int n = 5;
int p = 1;
while(n -- )
{
int t = x % 10;
x /= 10;
if(t) a.push_back(t * p);
p *= 10;
}
cout << a.size() << endl;
for(auto t : a)
cout << t << " ";
cout << endl;
}
int main()
{
freopen("test.in", "r", stdin);
ios::sync_with_stdio(false);
cin.tie(nullptr), cout.tie(nullptr);
int _t = 1;
cin >> _t;
while(_t -- ){
solve();
}
return 0;
}B. Same Parity Summands
给定两个整数n， k。构建一个k个奇偶性相同的数列，使得和是n。构造题，两种方式
如果前k-1个数写成1，最后留一个n-k+1也是奇数，就构造成功
如果前k-2个数为2，最后一个n - 2 * (k - 1) 也是偶数，就构造成功
否则就不能构造出来
#include <bits/stdc++.h>
using namespace std;
void solve(){
int n, k;
cin >> n >> k;
int n1 = n - (k - 1);
if(n1 > 0 && n1 % 2 == 1)
{
cout << "YES" << endl;
for(int i = 0; i < k - 1; i ++ )
cout << "1 ";
cout << n1 << endl;
return;
}
int n2 = n - 2 * (k - 1);
if(n2 > 0 && n2 % 2 == 0)
{
cout << "YES" << endl;
for(int i = 0; i < k - 1; i ++ )
cout << "2 ";
cout << n2 << endl;
return;
}
cout << "NO" << endl;
}
int main()
{
freopen("test.in", "r", stdin);
ios::sync_with_stdio(false);
cin.tie(nullptr), cout.tie(nullptr);
int _t = 1;
cin >> _t;
while(_t -- ){
solve();
}
return 0;
}C. K-th Not Divisible by n
题意很简单， 输出第 K 个不能被 n 整除的数
#include<bits/stdc++.h>
using namespace std;
void solve()
{
int n, k;
cin >> n >> k;
long long a = k / (n - 1);
long long b = k % (n - 1);
if(b == 0) b = -1;
cout << (long long)a * n + b << endl;
}
int main()
{
freopen("test.in", "r", stdin);
int tt;
cin >> tt;
while(tt -- )
{
solve();
}
return 0;
}D. Alice, Bob and Candies
模拟
#include<bits/stdc++.h>
using namespace std;
const int N = 1010;
int a[N];
void solve()
{
int n;
cin >> n;
for(int i = 0; i < n; i ++ )
cin >> a[i];
int ans1 = 0, ans2 = 0;
int res1 = 0, res2 = 0;
int l = 0, r = n - 1;
int cnt = 0;
while(l <= r)
{
if(res1 <= res2)
{
res1 = 0;
while(res1 <= res2 && l <= r)
{
res1 += a[l];
ans1 += a[l];
l ++ ;
}
cnt ++ ;
}
else if(res1 >= res2)
{
res2 = 0;
while(res2 <= res1 && l <= r)
{
res2 += a[r];
ans2 += a[r];
r -- ;
}
cnt ++ ;
}
}
cout << cnt << " " << ans1 << " " << ans2 << endl;
}
int main()
{
freopen("test.in", "r", stdin);
int tt;
cin >> tt;
while(tt -- )
{
solve();
}
return 0;
}E. Special Elements
定义一种特殊元素，只要有连续区间和等于它，它就是特殊元素，问有多少个特殊元素先用cnt记录每个数字出先的次数，然后n^2遍历所有区间，算出区间和t，ans += cnt[t];
#include<bits/stdc++.h>
using namespace std;
const int N = 10010;
typedef long long LL;
int s[N];
int cnt[N];
bool st[N];
void solve()
{
memset(st, 0, sizeof st);
memset(cnt, 0, sizeof cnt);
int n;
cin >> n;
for(int i = 1; i <= n; i ++ )
{
cin >> s[i];
cnt[s[i]] ++ ;
s[i] += s[i - 1];
}
LL ans = 0;
for(int len = 2; len <= n; len ++ )
for(int l = 1; l + len - 1 <= n; l ++ )
{
int r = l + len - 1;
LL t = s[r] - s[l - 1];
if(t < N && !st[t])
{
ans += cnt[t];
st[t] = true;
}
}
cout << ans << endl;
}
int main()
{
freopen("test.in", "r", stdin);
int tt;
cin >> tt;
while(tt -- )
{
solve();
}
return 0;
}F. Binary String Reconstruction
构造，特殊情况特判
#include <bits/stdc++.h>
using namespace std;
void solve(){
int a, b, c;
cin >> a >> b >> c;
bool f = false;
string s;
if(a == b && a == 0)
{
a = -1;
c ++ ;
}
if(b > 0 && b % 2 == 0)
{
b -- ;
f = true;
}
for(int i = 1; i <= a + 1; i ++ )
s += '0';
for(int i = 1; i <= b; i ++ )
{
s += (char)('0' + (i & 1));
}
for(int i = 1; i <= c; i ++ )
s += '1';
if(f) s += '0';
cout << s << "\n";
}
int main()
{
freopen("test.in", "r", stdin);
ios::sync_with_stdio(false);
cin.tie(nullptr), cout.tie(nullptr);
int _t = 1;
cin >> _t;
while(_t -- ){
solve();
}
return 0;
}G. Special Permutation
如果n小于4，肯定不存在剩下的先从大到小输出奇数，在输出4和2，在从6开始从小到大输出偶数
#include <bits/stdc++.h>
using namespace std;
void solve(){
int n;
cin >> n;
if(n < 4)
{
cout << -1 << endl;
return;
}
int t = n;
if(t % 2 == 0) t -- ;
for(int i = t; i > 0; i -= 2)
cout << i << " ";
cout << 4 << " " << 2 << " ";
for(int i = 6; i <= n; i += 2)
cout << i << " ";
cout << endl;
}
int main()
{
freopen("test.in", "r", stdin);
ios::sync_with_stdio(false);
cin.tie(nullptr), cout.tie(nullptr);
int _t = 1;
cin >> _t;
while(_t -- ){
solve();
}
return 0;
}
