AC修炼计划（AtCoder Regular Contest 180） A~C

A - ABA and BAB

A - ABA and BAB (atcoder.jp)

这道题我一开始想复杂了，一直在想怎么dp，没注意到其实是个很简单的规律题。

我们可以发现我们住需要统计一下类似ABABA这样不同字母相互交替的所有子段的长度，而每个字段的的情况有（长度+1）/2种，最后所有字段情况的乘积就是最终答案。

#pragma GCC optimize(3)  //O2优化开启
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
// const int mod=1e9+7;
const int MX=0x3f3f3f3f3f3f3f3f; 
//inline int read()                     //快读
//{
//    int xr=0,F=1; char cr;
//    while(cr=getchar(),cr<'0'||cr>'9') if(cr=='-') F=-1;
//    while(cr>='0'&&cr<='9')
//        xr=(xr<<3)+(xr<<1)+(cr^48),cr=getchar();
//    return xr*F;
//}
//void write(int x)                     //快写
//{
//    if(x<0) putchar('-'),x=-x;
//    if(x>9) write(x/10); putchar(x%10+'0');
//}
// 比 unordered_map 更快的哈希表
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
// const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();
// struct chash {
//     int operator()(int x) const { return x ^ RANDOM; }
// };
// typedef gp_hash_table<int, int, chash> hash_t;
constexpr ll mod = 1e9 + 7;                                //此处为自动取模的数
class modint{ll num;
public:modint(ll num = 0) :num(num % mod){}ll val() const {return num;}modint pow(ll other) {modint res(1), temp = *this;while(other) {if(other & 1) res = res * temp;temp = temp * temp;other >>= 1;}return res;}constexpr ll norm(ll num) const {if (num < 0) num += mod;if (num >= mod) num -= mod;return num;}modint inv(){ return pow(mod - 2); }modint operator+(modint other){ return modint(num + other.num); }modint operator-(){ return { -num }; }modint operator-(modint other){ return modint(-other + *this); }modint operator*(modint other){ return modint(num * other.num); }modint operator/(modint other){ return *this * other.inv(); }modint &operator*=(modint other) { num = num * other.num % mod; return *this; }modint &operator+=(modint other) { num = norm(num + other.num); return *this; }modint &operator-=(modint other) { num = norm(num - other.num); return *this; }modint &operator/=(modint other) { return *this *= other.inv(); }friend istream& operator>>(istream& is, modint& other){ is >> other.num; other.num %= mod; return is; }friend ostream& operator<<(ostream& os, modint other){ other.num = (other.num + mod) % mod; return os << other.num; }
};int n;
string s;
void icealsoheat(){cin>>n;cin>>s;s=" "+s;int res=1;modint ans=1;for(int i=2;i<=n;i++){if(s[i]!=s[i-1]){res++;}else{if(res>=3){ans*=(res+1)/2;}res=1;}}if(res>=3){ans*=(res+1)/2;}cout<<ans;
}
signed main(){ios::sync_with_stdio(false);          //int128不能用快读！！！！！！cin.tie();cout.tie();int _yq;_yq=1;// cin>>_yq;while(_yq--){icealsoheat();}
}
//
//⠀⠀⠀             ⠀⢸⣿⣿⣿⠀⣼⣿⣿⣦⡀
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⠀⠀⠀ ⠀⢸⣿⣿⡟⢰⣿⣿⣿⠟⠁
//⠀⠀⠀⠀⠀⠀⠀⢰⣿⠿⢿⣦⣀⠀⠘⠛⠛⠃⠸⠿⠟⣫⣴⣶⣾⡆
//⠀⠀⠀⠀⠀⠀⠀⠸⣿⡀⠀⠉⢿⣦⡀⠀⠀⠀⠀⠀⠀ ⠛⠿⠿⣿⠃
//⠀⠀⠀⠀⠀⠀⠀⠀⠙⢿⣦⠀⠀⠹⣿⣶⡾⠛⠛⢷⣦⣄⠀
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣧⠀⠀⠈⠉⣀⡀⠀ ⠀⠙⢿⡇
//⠀⠀⠀⠀⠀⠀⢀⣠⣴⡿⠟⠋⠀⠀⢠⣾⠟⠃⠀⠀⠀⢸⣿⡆
//⠀⠀⠀⢀⣠⣶⡿⠛⠉⠀⠀⠀⠀⠀⣾⡇⠀⠀⠀⠀⠀⢸⣿⠇
//⢀⣠⣾⠿⠛⠁⠀⠀⠀⠀⠀⠀⠀⢀⣼⣧⣀⠀⠀⠀⢀⣼⠇
//⠈⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⡿⠋⠙⠛⠛⠛⠛⠛⠁
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣾⡿⠋⠀
//⠀⠀⠀⠀⠀⠀⠀⠀⢾⠿⠋⠀
//

B - Improve Inversions

B - Improve Inversions (atcoder.jp)

这题确实不好想，但是get到点儿了就会觉得其实也不难。

我们需要尽可能的把逆序对最大化，从样例三我们可以发现，我们不妨确立左边一个要交换的下标i，然后找下标大于等于i+k，数值从大到小的找小于ai的数字，并依次与i的数字进行交换。为了尽可能减少替换的影响，我们按数值从小到大的次序去找这个下标i，从而参与上述的交换。因为我们是按从小到大的顺序的，所以小的数字参与交换并不会影响后面大的数字该交换的逆序对。

#pragma GCC optimize(3)  //O2优化开启
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
// const int mod=1e9+7;
const int MX=0x3f3f3f3f3f3f3f3f; 
//inline int read()                     //快读
//{
//    int xr=0,F=1; char cr;
//    while(cr=getchar(),cr<'0'||cr>'9') if(cr=='-') F=-1;
//    while(cr>='0'&&cr<='9')
//        xr=(xr<<3)+(xr<<1)+(cr^48),cr=getchar();
//    return xr*F;
//}
//void write(int x)                     //快写
//{
//    if(x<0) putchar('-'),x=-x;
//    if(x>9) write(x/10); putchar(x%10+'0');
//}
// 比 unordered_map 更快的哈希表
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
// const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();
// struct chash {
//     int operator()(int x) const { return x ^ RANDOM; }
// };
// typedef gp_hash_table<int, int, chash> hash_t;
// constexpr ll mod = 1e9 + 7;                                //此处为自动取模的数
// class modint{
//     ll num;
// public:
//     modint(ll num = 0) :num(num % mod){}//     ll val() const {
//         return num;
//     }//     modint pow(ll other) {
//         modint res(1), temp = *this;
//         while(other) {
//             if(other & 1) res = res * temp;
//             temp = temp * temp;
//             other >>= 1;
//         }
//         return res;
//     }//     constexpr ll norm(ll num) const {
//         if (num < 0) num += mod;
//         if (num >= mod) num -= mod;
//         return num;
//     }//     modint inv(){ return pow(mod - 2); }
//     modint operator+(modint other){ return modint(num + other.num); }
//     modint operator-(){ return { -num }; }
//     modint operator-(modint other){ return modint(-other + *this); }
//     modint operator*(modint other){ return modint(num * other.num); }
//     modint operator/(modint other){ return *this * other.inv(); }
//     modint &operator*=(modint other) { num = num * other.num % mod; return *this; }
//     modint &operator+=(modint other) { num = norm(num + other.num); return *this; }
//     modint &operator-=(modint other) { num = norm(num - other.num); return *this; }
//     modint &operator/=(modint other) { return *this *= other.inv(); }
//     friend istream& operator>>(istream& is, modint& other){ is >> other.num; other.num %= mod; return is; }
//     friend ostream& operator<<(ostream& os, modint other){ other.num = (other.num + mod) % mod; return os << other.num; }
// };int n,k;
int a[200005];
int p[200005];
vector<PII>ans;
void icealsoheat(){cin>>n>>k;int bns=0;for(int i=1;i<=n;i++)cin>>a[i],p[a[i]]=i;for(int i=1;i<=n;i++){int id=p[i];int x=i;for(int j=i-1;j>=1;j--){if(p[j]>=id+k){// cout<<p[x]<<":::"<<p[j]<<"\n";ans.push_back({p[x],p[j]});a[id]=j;a[p[j]]=x;swap(p[x],p[j]);x=j;}}}cout<<ans.size()<<"\n";for(auto [i,j]:ans){cout<<i<<" "<<j<<"\n";}// for(int i=1;i<=n;i++){//     cout<<a[i]<<" ";// }}
signed main(){ios::sync_with_stdio(false);          //int128不能用快读！！！！！！cin.tie();cout.tie();int _yq;_yq=1;// cin>>_yq;while(_yq--){icealsoheat();}
}
//
//⠀⠀⠀             ⠀⢸⣿⣿⣿⠀⣼⣿⣿⣦⡀
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⠀⠀⠀ ⠀⢸⣿⣿⡟⢰⣿⣿⣿⠟⠁
//⠀⠀⠀⠀⠀⠀⠀⢰⣿⠿⢿⣦⣀⠀⠘⠛⠛⠃⠸⠿⠟⣫⣴⣶⣾⡆
//⠀⠀⠀⠀⠀⠀⠀⠸⣿⡀⠀⠉⢿⣦⡀⠀⠀⠀⠀⠀⠀ ⠛⠿⠿⣿⠃
//⠀⠀⠀⠀⠀⠀⠀⠀⠙⢿⣦⠀⠀⠹⣿⣶⡾⠛⠛⢷⣦⣄⠀
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣧⠀⠀⠈⠉⣀⡀⠀ ⠀⠙⢿⡇
//⠀⠀⠀⠀⠀⠀⢀⣠⣴⡿⠟⠋⠀⠀⢠⣾⠟⠃⠀⠀⠀⢸⣿⡆
//⠀⠀⠀⢀⣠⣶⡿⠛⠉⠀⠀⠀⠀⠀⣾⡇⠀⠀⠀⠀⠀⢸⣿⠇
//⢀⣠⣾⠿⠛⠁⠀⠀⠀⠀⠀⠀⠀⢀⣼⣧⣀⠀⠀⠀⢀⣼⠇
//⠈⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⡿⠋⠙⠛⠛⠛⠛⠛⠁
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣾⡿⠋⠀
//⠀⠀⠀⠀⠀⠀⠀⠀⢾⠿⠋⠀
//

C - Subsequence and Prefix Sum

C - Subsequence and Prefix Sum (atcoder.jp)

一道非常巧妙的dp题，他的状态转移非常的奇妙。

我们考虑前i位的数字对后面数字的贡献值。可以分成两种情况。

1.第i位数字没有被选中

2.第i位数字被选中

当第i位数字被选中时，每一个位数i它所能合成的状态数字都对后面i+1到n的数字有相应的贡献。而这里面0的情况比较特殊，如果第i位的合成数字是0，其实不会改变下一个选中的数字。

这里面有一种情况比较特殊

例如1 -1 5 5 .........

这里我们会发现，我们选择1和-1后，选择第3个5和第4个5的情况是重复的，所以我们要想办法将它去重。

#pragma GCC optimize(3)  //O2优化开启
#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
// const int mod=1e9+7;
const int MX=0x3f3f3f3f3f3f3f3f; 
//inline int read()                     //快读
//{
//    int xr=0,F=1; char cr;
//    while(cr=getchar(),cr<'0'||cr>'9') if(cr=='-') F=-1;
//    while(cr>='0'&&cr<='9')
//        xr=(xr<<3)+(xr<<1)+(cr^48),cr=getchar();
//    return xr*F;
//}
//void write(int x)                     //快写
//{
//    if(x<0) putchar('-'),x=-x;
//    if(x>9) write(x/10); putchar(x%10+'0');
//}
// 比 unordered_map 更快的哈希表
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
// const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();
// struct chash {
//     int operator()(int x) const { return x ^ RANDOM; }
// };
// typedef gp_hash_table<int, int, chash> hash_t;
constexpr ll mod = 1e9 + 7;                                //此处为自动取模的数
class modint{ll num;
public:modint(ll num = 0) :num(num % mod){}ll val() const {return num;}modint pow(ll other) {modint res(1), temp = *this;while(other) {if(other & 1) res = res * temp;temp = temp * temp;other >>= 1;}return res;}constexpr ll norm(ll num) const {if (num < 0) num += mod;if (num >= mod) num -= mod;return num;}modint inv(){ return pow(mod - 2); }modint operator+(modint other){ return modint(num + other.num); }modint operator-(){ return { -num }; }modint operator-(modint other){ return modint(-other + *this); }modint operator*(modint other){ return modint(num * other.num); }modint operator/(modint other){ return *this * other.inv(); }modint &operator*=(modint other) { num = num * other.num % mod; return *this; }modint &operator+=(modint other) { num = norm(num + other.num); return *this; }modint &operator-=(modint other) { num = norm(num - other.num); return *this; }modint &operator/=(modint other) { return *this *= other.inv(); }friend istream& operator>>(istream& is, modint& other){ is >> other.num; other.num %= mod; return is; }friend ostream& operator<<(ostream& os, modint other){ other.num = (other.num + mod) % mod; return os << other.num; }
};int n,k;
int a[500005];
modint dp[105][5005];
modint sum[5005];
void icealsoheat(){cin>>n;for(int i=0;i<=20;i++)if(i!=10)sum[i]=1;for(int i=1;i<=n;i++)cin>>a[i];modint ans=1;for(int i=0;i<n;i++){dp[i][a[i]+1000]=dp[i][a[i]+1000]+sum[a[i]+10];sum[a[i]+10]=0;for(int j=0;j<=2000;j++){if(j==1000)continue;for(int o=i+1;o<=n;o++){// if(j+a[o]<0)cout<<"+++\n";if(j+a[o]<0)continue;dp[o][j+a[o]]+=dp[i][j];ans+=dp[i][j];}}for(int j=0;j<=20;j++){if(j!=10)sum[j]+=dp[i][1000];}}cout<<dp[2][1]<<"+++\n";cout<<ans;}
signed main(){ios::sync_with_stdio(false);          //int128不能用快读！！！！！！cin.tie();cout.tie();int _yq;_yq=1;// cin>>_yq;while(_yq--){icealsoheat();}
}
//
//⠀⠀⠀             ⠀⢸⣿⣿⣿⠀⣼⣿⣿⣦⡀
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⠀⠀⠀ ⠀⢸⣿⣿⡟⢰⣿⣿⣿⠟⠁
//⠀⠀⠀⠀⠀⠀⠀⢰⣿⠿⢿⣦⣀⠀⠘⠛⠛⠃⠸⠿⠟⣫⣴⣶⣾⡆
//⠀⠀⠀⠀⠀⠀⠀⠸⣿⡀⠀⠉⢿⣦⡀⠀⠀⠀⠀⠀⠀ ⠛⠿⠿⣿⠃
//⠀⠀⠀⠀⠀⠀⠀⠀⠙⢿⣦⠀⠀⠹⣿⣶⡾⠛⠛⢷⣦⣄⠀
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣧⠀⠀⠈⠉⣀⡀⠀ ⠀⠙⢿⡇
//⠀⠀⠀⠀⠀⠀⢀⣠⣴⡿⠟⠋⠀⠀⢠⣾⠟⠃⠀⠀⠀⢸⣿⡆
//⠀⠀⠀⢀⣠⣶⡿⠛⠉⠀⠀⠀⠀⠀⣾⡇⠀⠀⠀⠀⠀⢸⣿⠇
//⢀⣠⣾⠿⠛⠁⠀⠀⠀⠀⠀⠀⠀⢀⣼⣧⣀⠀⠀⠀⢀⣼⠇
//⠈⠋⠁⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⡿⠋⠙⠛⠛⠛⠛⠛⠁
//⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⣾⡿⠋⠀
//⠀⠀⠀⠀⠀⠀⠀⠀⢾⠿⠋⠀
//