A. Favorite Sequence

思路：照着模拟即可。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll a[maxn];

ll b[maxn];

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        ll n = read();

        rep(i,1,n) a[i] = read();

        int L = 1, R= n;

        int pos = 0;

        while(L<=R)

        {

            if(pos&1)

            b[++pos] = a[R--];

            else b[++pos] = a[L++];

        }

        rep(i,1,n) cout<<b[i]<<' '; cout<<endl;

    }

    return 0;

}

B. Last Year's Substring

思路：题意说要去掉一个子串后剩下2020。相当于切割字符串，去掉中间一部分，留下头尾。

所以只用分类讨论一下头尾各自的长度，看看能不能凑成2020。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

string p = "20";

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        ll n = read();

        string s;

        cin>>s;

        int flag = 0;

        if(s.size()>=4&&s[0]=='2'&&s[1]=='0'&&s[n-2]=='2'&&s[n-1]=='0') flag = 1;

        if(s.size()>=4&&s[n-4]=='2'&&s[n-3]=='0'&&s[n-2]=='2'&&s[n-1]=='0') flag = 1;

        if(s.size()>=4&&s[0]=='2'&&s[1]=='0'&&s[2]=='2'&&s[3]=='0') flag = 1;

        if(s.size()>=4&&s[0]=='2'&&s[n-3]=='0'&&s[n-2]=='2'&&s[n-1]=='0') flag = 1;

        if(s.size()>=4&&s[0]=='2'&&s[1]=='0'&&s[n-2]=='2'&&s[n-1]=='0') flag = 1;

        if(s.size()>=4&&s[0]=='2'&&s[1]=='0'&&s[2]=='2'&&s[n-1]=='0') flag = 1;

        puts(flag?"YES":"NO");

    }

    return 0;

}

C. Unique Number

思路：贪心，把数字大的填到低位，既能保证数位长度尽可能小又能使得整体的数值更小。所以从9往前填数看看能不能填到即可。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        ll n = read();

        ll cur = 9;

        string s;

        int flag = 1;

        while(n&&cur>0)

        {

            ll mi = min(cur,n);

            if(mi > cur)

            {

                flag = 0;

                break;

            }

            cur = mi;

            s += cur + '0';

            n -= cur;

            cur --;

        }

        if(n) flag = 0;

        if(!flag)

        {

            cout<<-1<<endl;

            continue;

        }

        for(int i=0, j=s.size()-1; i<=j; i++, j--) swap(s[i],s[j]);

        cout<<s<<endl;

    }

    return 0;

}

D. Add to Neighbour and Remove

思路：考虑合并出来的第一块。第一块肯定是到某个数的前缀和，我们就枚举一遍第一块的值sum[i]，往后检验以这个sum[i]作为最后那个相等的数是否合法，并且每次更新步数最小值即可。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll a[maxn];

ll sum[maxn];

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        ll n = read();

        rep(i,1,n) a[i] = read(), sum[i] = sum[i-1] + a[i];

        ll mi = inf;

        rep(i,1,n)

        {

            ll cur = sum[i];

            ll step = i-1;

            int flag = 1;

            ll p = i;

            int to = i;

            rep(j,i+1,n)

            {

                if(sum[j] - sum[p]==cur)

                {

                    step += (j-p-1);

                    p = j;

                    to = j;

                }

                else if(sum[j] - sum[p] > cur)

                {

                    flag = 0;

                    break;

                }

            }

            if(flag&&to==n) mi = min(mi, step);

        }

        cout<<mi<<endl;

    }

    return 0;

}

E. Close Tuples

思路：我做这两题的方法一致，直接拿E2讲了。

先对a数组排序，然后我们就可以枚举a[i],每次用lower_bound找到第一个大于等于a[i]-k的位置id，然后在id~i-1中选m-1个数的方案就是以a[i]结尾的m元组产生的贡献。累加即可。

view code 
#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define endl '\n'

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 2e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll a[maxn];

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        ll n = read(), m = read(), k = read();

        rep(i,1,n) a[i] = read();

        sort(a+1,a+1+n);

        ll ans = 0;

        rep(i,m,n)

        {

            int id = lower_bound(a+1,a+1+n,a[i]-k) - a;

            if(id<i&&i-id>=m-1)

            {

                ll num = (i-id);  // num units choose m

                ll chosen = min(m-1,num-(m-1));

                ll up = 1;

                ll down = 1;

                for(int j=num, l=1; l<=chosen; j--, l++)

                {

                    up = (up%mod * j%mod) %mod;

                    down = (l%mod * down%mod)%mod;

                }

                ll dd = inv(down,mod);

                ans = (ans%mod + (up*dd)%mod ) %mod;

            }

            else if(m==1) ans += 1;

        }

        cout<<ans<<endl;

    }

    return 0;

}

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