目录
A. Only Pluses
当三个数更接近与同一个数时，三个的乘积就是最大的。
#include<iostream>
#include<algorithm>
#include<queue>
#include<math.h>
#include<cstring>
#include<map>
#define ll long long
#define TEST int T;cin>>T;while(T--)
#define lowbit(x) x&(-x)
using namespace std;
const int N = 2000100;
const int M = 1e8 + 7;
int base1 = 131, base2 = 13311;
void solve() {
vector<ll>q;
for(int i=1,x;i<=3;i++)
{
cin>>x;
q.push_back(x);
}
for(int i=1;i<=5;i++)
{
sort(q.begin(),q.end());
q[0]++;
}
cout<<q[0]*q[1]*q[2]<<"\n";
}
int main() {
TEST
solve();
return 0;
}B. Angry Monk
先找到数组的最大数，不需要对它处理，只需将分出的1和它合并即可，想将a[i]分成a[i]个1需要a[i]-1步，将a[i]个1合并需要a[i]步。
#include<iostream>
#include<algorithm>
#include<queue>
#include<math.h>
#include<cstring>
#include<map>
#define ll long long
#define TEST int T;cin>>T;while(T--)
#define lowbit(x) x&(-x)
using namespace std;
const int N = 200100;
const int M = 1e8 + 7;
int base1 = 131, base2 = 13311;
ll a[N];
void solve() {
ll n,k;
cin>>n>>k;
ll ans=0,p=-1;
for(int i=1;i<=k;i++)
{
cin>>a[i];
if(a[i]>ans)
{
p=i;
ans=a[i];
}
}
ll sum=0;
for(int i=1;i<=k;i++)
{
if(i!=p)
{
if(a[i]==1) sum++;
else sum+=a[i]-1,sum+=a[i];
}
}
cout<<sum<<"\n";
}
int main() {
TEST
solve();
return 0;
}C. Gorilla and Permutation
我们需要最大化全部f(x)的和，最小化全部g(x)的和。那么我们就需要让大于k的数提早出现并且最大，让小于m的数最晚出现。其他数字顺便排。
#include<iostream>
#include<algorithm>
#include<queue>
#include<math.h>
#include<cstring>
#include<map>
#define ll long long
#define TEST int T;cin>>T;while(T--)
#define lowbit(x) x&(-x)
using namespace std;
const int N = 200100;
const int M = 1e8 + 7;
int base1 = 131, base2 = 13311;
void solve() {
ll n,m,k;
cin>>n>>m>>k;
for(int i=n;i>=k;i--) cout<<i<<" ";
for(int i=k-1;i>=m+1;i--) cout<<i<<" ";
for(int i=1;i<=m;i++) cout<<i<<" ";
cout<<"\n";
}
int main() {
TEST
solve();
return 0;
}D. Test of Love
模拟+贪心
#include<iostream>
#include<algorithm>
#include<queue>
#include<math.h>
#include<cstring>
#include<map>
#define ll long long
#define TEST int T;cin>>T;while(T--)
#define lowbit(x) x&(-x)
using namespace std;
const int N = 200100;
const int M = 1e8 + 7;
int base1 = 131, base2 = 13311;
void solve()
{
ll ans=0;
ll n,m,k;
string s;
cin >> n >> m >> k;
cin >> s;
s = " " + s;
if (m > n) {
cout << "YES" << endl;
return;
}
else {
ans = m;
for (int i = 1; i <= n; i++) {
if (ans <= 0) {
cout << "NO" << endl;
return;
}
if (s[i] == 'L')
ans = m;
if (s[i] == 'W') {
if (k > 0) {
if (ans > 1)
ans--;
else {
ans = 1;
k--;
}
}
else
ans--;
}
if (s[i] == 'C')
ans--;
}
}
if (ans > 0)
cout << "YES" << "\n";
else
cout << "NO" << "\n";
}
int main() {
TEST
solve();
return 0;
}E. Novice's Mistake
我们观察ab和n的范围，发现a*n-b最大不超过1000000，所以我们可以暴力求解答案，将求解过程分为n为1位数和二位数，三位数是0。
#include<iostream>
#include<algorithm>
#include<queue>
#include<math.h>
#include<cstring>
#include<map>
#define ll long long
#define TEST int T;cin>>T;while(T--)
#define lowbit(x) x&(-x)
using namespace std;
const int N = 200100;
const int M = 1e8 + 7;
int base1 = 131, base2 = 13311;
ll quick(ll base, ll power) {
ll ans = 1;
while (power) {
if (power & 1) {
ans = (ans * base);
}
base = base * base;
power >>= 1;
}
return ans;
}
void solve() {
ll n;
cin >> n;
if (n == 100) {
cout << 0 << "\n";
return ;
}
vector<ll>ans_A, ans_B;
if (n < 10) { //一位数
for (int a = 1; a <= 10000; a++) {
for (int b = max(1, a - 6); b < a; b++) {
ll re = n * (quick(10, a - b) - 1) / 9;
if (a * n - b == re) {
ans_A.push_back(a);
ans_B.push_back(b);
}
}
}
} else { //两位数
for (int a = 1; a <= 10000; a++) {
for (int b = max(1, 2 * a - 6); b < 2 * a ; b++) {
if (b & 1) {
ll cnt = 1;
ll re = 0;
for (int i = 0; i < (2 * a - b) / 2; i++) {
re += n * quick(10, cnt);
cnt += 2;
}
re += n / 10;
if (a * n - b == re) {
ans_A.push_back(a);
ans_B.push_back(b);
}
} else {//偶数可以正常处理
ll cnt = 0;
ll re = 0;
for (int i = 0; i < (2 * a - b) / 2; i++) {
re += n * quick(10, cnt);
cnt += 2;
}
if (a * n - b == re) {
ans_A.push_back(a);
ans_B.push_back(b);
}
}
}
}
}
cout << ans_A.size() << "\n";
for (int i = 0; i < ans_A.size(); i++) {
cout << ans_A[i] << " " << ans_B[i] << "\n";
}
}
int main() {
TEST
solve();
return 0;
}
