ACM日常训练日记——7.26（线性dp+cf div2）
- Atcoder训练
- Powerful Discount Tickets
我们只需要动态维护使最大的值变小即可，这里我采用multiset去记录，有相同元素存在，也可以采用优先队列去维护
- Powerful Discount Tickets
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
ll n,m;
ll v[100010];
multiset<ll>st;
multiset<ll>::iterator ip;
int main(){
cin>>n>>m;
for(int i=1;i<=n;i++)cin>>v[i],st.insert(v[i]);
for(int i=1;i<=m;i++){
ip = st.end();
st.erase(--ip);
st.insert(*ip/2);
}
ll ans=0;
for(auto it=st.begin();it!=st.end();it++){
ans+=*it;
}
cout<<ans;
}
- 动态规划专项训练
1.方块与收纳盒
这道题很像走楼梯问题，是一个经典的线性dp，楼梯每次只能走一楼和二楼，所以dp[i]=dp[i-1]+dp[i-2]的方案数
这里也一样
#include <bits/stdc++.h>
using namespace std;
long long dp[1001];
int main(){
int n;
cin>>n;
dp[1]=1;
dp[2]=2;
for(int i=3;i<=80;i++)dp[i]=dp[i-1]+dp[i-2];
while(n--){
long long t;
cin>>t;
cout<<dp[t]<<'\n';
}
}
2.舔狗舔到最后一无所有
dp[i]代表前i天有多少种方案，分两种情况：(把问题转化成i和i-1不同，i和i-2不同)
1、第i天的饭菜与第i-2天的饭菜不同，那么前i天有dp[i-2]x2种方案2、第i天的饭菜与第i-1天的饭菜不同，那么前i天有dp[i-1]x2种方案
因此前i天共计dp[i-1]*2+dp[i-2]x2,在这中包含都不同的情况
#include <bits/stdc++.h>
#define size 100100
const int MOD=1e9+7;
using namespace std;
long long dp[size],n,m;
int main(){
dp[1]=3,dp[2]=9;
for (int i=3;i<size;i++){
dp[i]=(dp[i-1]*2%MOD+dp[i-2]*2%MOD)%MOD;
}
scanf("%lld",&m);
while (m--){
scanf("%lld",&n);
printf("%lld\n",dp[n]);
}
system("pause");
return 0;
}
- Codeforcesz(Codeforces Round 962 (Div. 3))
1.Left Right Operation
送分题，如果能整除4，答案除4，不是的话加上1
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
int main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int n;
cin>>n;
while(n--){
ll t;
cin>>t;
if(t<=4)cout<<1<<'\n';
else{
if(t%4==0)cout<<t/4<<'\n';
else cout<<t/4+1<<'\n';
}
}
}
2. Scale
继续送分，画图，它的图是以k个为单位画的，每次画加k即可
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int t;
cin >> t;
while (t--) {
int n, k;
cin >> n >> k;
vector<string> grid(n);
for (int i = 0; i < n; ++i) {
cin >> grid[i];
}
int reduced_n = n / k;
vector<string> reduced_grid(reduced_n, string(reduced_n, '0'));
for (int i = 0; i < reduced_n; ++i) {
for (int j = 0; j < reduced_n; ++j) {
int start_row = i * k;
int start_col = j * k;
reduced_grid[i][j] = grid[start_row][start_col];
}
}
for (int i = 0; i < reduced_n; ++i) {
cout << reduced_grid[i] << '\n';
}
}
return 0;
}
3.Sort
用二维前缀和来记录两个区间的字母数量和类型，然后进行判断大小，返回差值即可
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
vector<vector<int>> preprocess(const string& s) {
int n = s.size();
vector<vector<int>> prefix(26, vector<int>(n + 1, 0));
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < 26; ++j) {
prefix[j][i] = prefix[j][i - 1];
}
prefix[s[i - 1] - 'a'][i]++;
}
return prefix;
}
int mi(const vector<vector<int>>& prefix_a, const vector<vector<int>>& prefix_b, int l, int r) {
int ans = 0;
for (int i = 0; i < 26; ++i) {
int ca = prefix_a[i][r] - prefix_a[i][l - 1];
int cb = prefix_b[i][r] - prefix_b[i][l - 1]; // 修正这里的错误
if (ca > cb) {
ans += (ca - cb);
}
}
return ans;
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t;
cin >> t;
while (t--) {
int n, q;
cin >> n >> q;
string a, b;
cin >> a >> b;
vector<vector<int>> prefix_a = preprocess(a);
vector<vector<int>> prefix_b = preprocess(b);
for (int i = 0; i < q; ++i) {
int l, r;
cin >> l >> r;
cout << mi(prefix_a, prefix_b, l, r) << '\n';
}
}
return 0;
}
4.Fun
暴力加上减枝即可
#include <bits/stdc++.h>
using namespace std;
long long c(long long n, long long x) {
long long count = 0;
for (long long a = 1; a < x - 1; ++a) {
for (long long b = 1; b < x - a; ++b) {
long long ab = a * b;
if (ab >= n) break; // 如果 ab 已经大于等于 n，则不需要继续计算
long long c1 = (a + b > 0) ? (n - ab) / (a + b) : 0;
long long c2 = x - a - b;
long long c_max = min(c1, c2);
if (c_max > 0) {
count += c_max;
}
}
}
return count;
}
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int t;
cin >> t;
vector<long long> results;
for (int i = 0; i < t; ++i) {
long long n, x;
cin >> n >> x;
results.push_back(c(n, x));
}
for (long long result : results) {
cout << result << '\n';
}
return 0;
}
