Codeforces Round #725 (Div. 3) A-D，F题解
A. Stone Game
思路：总共3种情况，都从最左端被拿走，都从最右端被拿走，左侧的从最左端被拿走且右侧的从最右端被拿走，取最小值即可
代码：
//CF-725.A
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
typedef long long ll;
typedef pair<int, int>P;
typedef vector<int>vec;
#define INF 0x3f3f3f3f
const double EPS = 1e-18;
const int MOD = 1e9 + 7;
const int maxn = 8e4 + 1;
int T;
int N, A[maxn];
void solve()
{
int maxi = -1, mini = -1, maxa = -1, mina = INF;
for (int i = 0; i < N; i++)
{
if (A[i] < mina)
{
mina = A[i];
mini = i;
}
if (A[i] > maxa)
{
maxa = A[i];
maxi = i;
}
}
if (mini < maxi)
{
int ans[3];
ans[0] = mini + 1 + N - maxi;
ans[1] = max(mini, maxi) + 1;
ans[2] = N - min(mini, maxi);
int res = INF;
for (int i = 0; i < 3; i++)
res = min(res, ans[i]);
cout << res << endl;
return;
}
else
{
int ans[3];
ans[0] = maxi + 1 + N - mini;
ans[1] = max(mini, maxi) + 1;
ans[2] = N - min(mini, maxi);
int res = INF;
for (int i = 0; i < 3; i++)
res = min(res, ans[i]);
cout << res << endl;
return;
}
}
int main()
{
IOS;
cin >> T;
while (T--)
{
cin >> N;
for (int i = 0; i < N; i++)
cin >> A[i];
solve();
}
return 0;
}B. Friends and Candies
思路：只能是从糖多的人那里把糖分给糖少的，所以只要求出糖数大于平均的人数即可，当糖数不能被人数整除时输出-1。
代码：
//CF-725.B
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
typedef long long ll;
typedef pair<int, int>P;
typedef vector<int>vec;
#define INF 0x3f3f3f3f
const double EPS = 1e-18;
const int MOD = 1e9 + 7;
const int maxn = 2e5 + 1;
int T;
ll N, A[maxn];
void solve()
{
int sum = 0, ans = 0;
int avg;
for (int i = 0; i < N; i++)
sum += A[i];
if (sum % N)
{
cout << -1 << endl;
return;
}
avg = sum / N;
for (int i = 0; i < N; i++)
{
if (A[i] > avg)
ans++;
}
cout << ans << endl;
}
int main()
{
IOS;
cin >> T;
while (T--)
{
cin >> N;
for (int i = 0; i < N; i++)
cin >> A[i];
solve();
}
return 0;
}C. Number of Pairs
又是因为没开longlongWA了半天。。。
思路：分治法，不断将数列分为左右两个数列，完全在其中一个数列中的数对数量可以递归求得。合并的同时将数列排序，对于跨两个数列的数对，只需枚举左边数列中的数，在右边找加上它之后处于区间内的数的个数即可。
代码：
//CF-725.C
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
typedef long long ll;
typedef pair<int, int>P;
typedef vector<int>vec;
#define INF 0x3f3f3f3f
const double EPS = 1e-18;
const int MOD = 1e9 + 7;
const int maxn = 1e9 + 1;
int T;
ll n, l, r;
vec A;
ll merge_solve(vec& a)
{
ll ans = 0;
ll d = a.size();
if (d <= 1)
return 0;
vec b(a.begin(), a.begin() + d / 2);
vec c(a.begin() + d / 2, a.end());
ans += merge_solve(b);
ans += merge_solve(c);
ll ai = 0, bi = 0, ci = 0;
while (ai < d)
{
if (bi < b.size() && (ci == c.size() || b[bi] < c[ci]))
{
if (b[bi] <= r)//只有小于右边界才可能可以
{
int hi = r - b[bi], lo = l - b[bi] - 1;
ans += upper_bound(c.begin(), c.end(), hi) - upper_bound(c.begin(), c.end(), lo);
}
a[ai++] = b[bi++];
}
else
a[ai++] = c[ci++];
}
return ans;
}
void solve()
{
cout << merge_solve(A) << endl;
}
int main()
{
IOS;
cin >> T;
while (T--)
{
A.clear();
cin >> n >> l >> r;
for (int i = 0; i < n; i++)
{
int num;
cin >> num;
A.push_back(num);
}
solve();
}
return 0;
}D. Another Problem About Dividing Numbers
思路：当K==1时，如果A和B不相等并且其中一个能整除另外一个时才YES，否则NO。当K!=1时，计算操作次数的上限，当把每个数都除到1的时候操作次数最多，所以直接统计每个数的质因数个数即可，两个数质因数的和M就是允许操作的最大次数。小于这个次数的操作显然都可以完成。如果K<=M，就YES，否则NO。
代码：
//CF-725.D
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
typedef long long ll;
typedef pair<int, int>P;
typedef vector<int>vec;
#define Y "YES"
#define N "NO"
#define INF 0x3f3f3f3f
const double EPS = 1e-18;
const int MOD = 1e9 + 7;
const int maxn = 1e9 + 1;
int T, A, B, K;
vec prime;
bool isprime(int n)
{
if (n == 0 || n == 1)
return false;
for (int i = 2; i <= sqrt(n); i++)
{
if (!(n % i))
return false;
}
return true;
}
void solve()
{
int numa = 0, numb = 0, index = 0;
if (K == 1)
{
if ((A != B) && (A % B == 0 || B % A == 0))
{
cout << Y << endl;
return;
}
else
cout << N << endl;
}
else
{
while (A != 1 && index < prime.size())
{
if (!(A % prime[index]))
{
A /= prime[index];
numa++;
}
else
index++;
}
if (A != 1)
numa++;
index = 0;
while (B != 1 && index < prime.size())
{
if (!(B % prime[index]))
{
B /= prime[index];
numb++;
}
else
index++;
}
if (B != 1)
numb++;
int times = numa + numb;
cout << (times >= K ? Y : N) << endl;
}
}
int main()
{
IOS;
for (int i = 0; i <= 1e5 + 7; i++)
{
if (isprime(i))
prime.push_back(i);
}
cin >> T;
while (T--)
{
cin >> A >> B >> K;
solve();
}
return 0;
}F. Interesting Function
没想到F题居然这么简单，快结束的时候看见一堆人过了才赶紧去做的。。。
思路：每加1就会有1个数字改变，如果进位了，每进1位就额外有1个数字改变，对于从a加到b，可以看作是从1加到b再减去从1加到a，进位的次数直接分别除以10，100……即可。
代码：
//CF-725.F
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
typedef long long ll;
typedef pair<int, int>P;
typedef vector<int>vec;
#define Y "YES"
#define N "NO"
#define INF 0x3f3f3f3f
const double EPS = 1e-18;
const int MOD = 1e9 + 7;
const int maxn = 1e9 + 1;
ll T, L, R;
ll A[10] = { 1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000 };
void solve()
{
ll ansr = 0, ansl = 0;
for (int i = 0; i < 10; i++)
{
ansr += R / A[i];
ansl += L / A[i];
}
cout << ansr - ansl << endl;
}
int main()
{
IOS;
cin >> T;
while (T--)
{
cin >> L >> R;
solve();
}
return 0;
}
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