在学校想打一场cf真是不容易.昨晚卡在E离散化的边界处理(也可能是思路错了)那里卡的很久,一直过不去就先睡觉去了.
这场的题目偏易.大多都可暴力乱搞过去.
A. Yet Another Two Integers Problem
题目大意:给两个数,每次可以减去1-10,问最少多少次A==B.
大概就是大数-10接近小数,如果%10的余数 大数大于小数的话,就要增加一次.

#pragma GCC optimize(2)
#define LL long long
#define pq priority_queue
#define ULL unsigned long long
#define pb push_back
#define mem(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define fir(i,a,b) for(int i=a;i<=(int)b;++i)
#define afir(i,a,b) for(int i=(int)a;i>=b;--i)
#define ft first
#define vi vector<int>
#define sd second
#define ALL(a) a.begin(),a.end()
#define bug puts("-------")
#define mpr(a,b) make_pair(a,b)
#include <bits/stdc++.h>
 
using namespace std;
const int N = 2e5+10;
 
inline void read(int &a){
    
    
    int x = 0,f=1;char ch = getchar();
    while(ch<'0'||ch>'9'){
    
    if(ch=='-')f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){
    
    x=x*10+ch-'0';ch=getchar();}
    a = x*f;
}
 
int solve(int a,int b){
    
    
    if(a < b) return solve(b,a);
    int ans = a/10-b/10;
    if(a%10>b%10) ans++;
    return ans;
}
int main(){
    
    
    int t;
    cin >> t;
    while(t--){
    
    
        int a,b;
        cin >> a >> b;
        cout << solve(a,b) << endl;
    }
    
    return 0;
}   

B. Minimum Product
题目大意:给a,b,x,y,n.a和b可以进行-1操作.这个操作次数不能超过n,而且a不能小于x,b不能小于y.问a×b能达到的最小值.
当时就乱搞了一下,要么先减a,要么先减b.贪心就过去了.毕竟数据范围不允许我们暴力.

#pragma GCC optimize(2)
#define LL long long
#define pq priority_queue
#define ULL unsigned long long
#define pb push_back
#define mem(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define fir(i,a,b) for(int i=a;i<=(int)b;++i)
#define afir(i,a,b) for(int i=(int)a;i>=b;--i)
#define ft first
#define vi vector<int>
#define sd second
#define ALL(a) a.begin(),a.end()
#define bug puts("-------")
#define mpr(a,b) make_pair(a,b)
#include <bits/stdc++.h>
 
using namespace std;
const int N = 2e5+10;
 
inline void read(int &a){
    
    
    int x = 0,f=1;char ch = getchar();
    while(ch<'0'||ch>'9'){
    
    if(ch=='-')f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){
    
    x=x*10+ch-'0';ch=getchar();}
    a = x*f;
}
 
LL solve(LL a,LL b,LL x,LL y,LL n){
    
    
    LL num = min(n,b-y);
    b -= num;
    n -= num;
    a -= min(n,a-x);
    return a*b;
}
int main(){
    
    
    int t;
    cin >> t;
    while(t--){
    
    
        LL a,b,x,y,n;
        cin >> a >> b >> x >> y >> n;
        cout << min(solve(a,b,x,y,n),solve(b,a,y,x,n)) << endl;
    }
    
    return 0;
}    

C. Yet Another Array Restoration
题目大意:给一个n,x,y 构造一个长度为n的等差数列.其中有x和y,要求输出能构造的所有方案中数列中最大值最小的那一个方案.
数据范围很小,暴力枚举公差,x的位置.y的位置就可以了.复杂度n^3.

#pragma GCC optimize(2)
#define LL long long
#define pq priority_queue
#define ULL unsigned long long
#define pb push_back
#define mem(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define fir(i,a,b) for(int i=a;i<=(int)b;++i)
#define afir(i,a,b) for(int i=(int)a;i>=b;--i)
#define ft first
#define vi vector<int>
#define sd second
#define ALL(a) a.begin(),a.end()
#define bug puts("-------")
#define mpr(a,b) make_pair(a,b)
#include <bits/stdc++.h>
 
using namespace std;
const int N = 2e5+10;
 
inline void read(int &a){
    
    
    int x = 0,f=1;char ch = getchar();
    while(ch<'0'||ch>'9'){
    
    if(ch=='-')f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){
    
    x=x*10+ch-'0';ch=getchar();}
    a = x*f;
}
 
int main(){
    
    
    int t,n,x,y;
    cin >> t;
    while(t--){
    
    
        cin >> n >> x >> y;
        int rd,rx,ry,mx=1e9;
        fir(d,1,50){
    
    
            afir(ypos,n,2){
    
    
                    afir(xpos,ypos-1,1){
    
    
                    if((ypos-xpos)*d == y-x && x-d*(xpos-1) > 0){
    
    
                        if((n-ypos)*d+y < mx){
    
    
                            mx = (n-ypos)*d+y;
                            rd = d;
                            rx = xpos;
                            ry = ypos;
                        }
                    }
                }
            }
        }   
        int sta = x-rd*(rx-1); 
        fir(i,1,n){
    
    
            cout << sta << " ";
            sta += rd;
        }   
        puts("");
    }
    return 0;
}    

D. Decrease the Sum of Digits
题目大意:给一个数,求最少加多少次1能让位数和<s.
每次让最后一个不是0的位数加到0然后每次都检查一下是否<s就可以了.

#pragma GCC optimize(2)
#define LL long long
#define pq priority_queue
#define ULL unsigned long long
#define pb push_back
#define mem(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define fir(i,a,b) for(int i=a;i<=(int)b;++i)
#define afir(i,a,b) for(int i=(int)a;i>=b;--i)
#define ft first
#define vi vector<int>
#define sd second
#define ALL(a) a.begin(),a.end()
#define bug puts("-------")
#define mpr(a,b) make_pair(a,b)
#include <bits/stdc++.h>
 
using namespace std;
const int N = 2e5+10;
 
inline void read(int &a){
    
    
    int x = 0,f=1;char ch = getchar();
    while(ch<'0'||ch>'9'){
    
    if(ch=='-')f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){
    
    x=x*10+ch-'0';ch=getchar();}
    a = x*f;
}
 
int calc(LL x){
    
    
    int res = 0;
    while(x){
    
    
        res += x%10;
        x/=10;
    }
    return res;
}
LL pre[20];
int main(){
    
    
    int t;
    cin >> t;
    pre[0] = 1;
    fir(i,1,18) pre[i] = pre[i-1]*10;
    while(t--){
    
    
        LL n;int s;
        cin >> n >> s;
        LL ans = 0;
        while(calc(n) > s){
    
    
            LL nn = n;
            int ct = 0;
            while(nn && !(nn%10)){
    
    
                ct++;
                nn/=10;
            }
            ans += 1LL*(10-nn%10)*pre[ct];
            n += 1LL*(10-nn%10)*pre[ct];
        }
        cout << ans << endl;
    }
    
    return 0;
}    

E. Two Platforms
题目大意(如果我没有理解错的话):给n个点的坐标,和两个长度为k的平板.问平板最多能覆盖多少个点(这个平板可以视为长度无限).
我的思路是要么连在一起成为一个k-2k的平板或者分开成两段.
那么就是一个很经典的单调队列问题,分两种情况写就好了.然而昨天调了半个多小时都没有调过去.
WA代码

#pragma GCC optimize(2)
#define LL long long
#define pq priority_queue
#define ULL unsigned long long
#define pb push_back
#define mem(a,x) memset(a,x,sizeof a)
#define pii pair<int,int>
#define fir(i,a,b) for(int i=a;i<=(int)b;++i)
#define afir(i,a,b) for(int i=(int)a;i>=b;--i)
#define ft first
#define vi vector<int>
#define sd second
#define ALL(a) a.begin(),a.end()
#define bug puts("-------")
#define mpr(a,b) make_pair(a,b)
#define L 2*i
#define R 2*i+1
#include <bits/stdc++.h>
 
using namespace std;
const int N = 2e5+10;
 
inline void read(int &a){
    
    
    int x = 0,f=1;char ch = getchar();
    while(ch<'0'||ch>'9'){
    
    if(ch=='-')f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){
    
    x=x*10+ch-'0';ch=getchar();}
    a = x*f;
}
int x[N],y[N],num[N],tot;
int q[N];
vi all;
int getpos(int x){
    
    
    return lower_bound(ALL(all),x) - all.begin();
}
int get_pos(int x){
    
    
    return upper_bound(ALL(all),x) - all.begin();
}
int main(){
    
    
    int t;
    cin >> t;
    while(t--){
    
    
        int n,k;
        cin >> n >> k;
        all.clear();
        mem(num,0);
        mem(q,0);
        all.pb(0);
        fir(i,1,n){
    
    
            cin >> x[i];
            all.pb(x[i]);
        }
        fir(i,1,n)
            cin >> y[i],all.pb(y[i]);
        sort(ALL(all));
        all.erase(unique(ALL(all)),all.end());
        tot = all.size()-1;
        fir(i,1,n)
            num[getpos(x[i])]++;
        int l=1,r=0;
        int ans = 1;
        fir(i,1,tot){
    
    
            num[i] += num[i-1];
            while(l <= r && all[i] - all[q[l]] >= 2*(k+1)) l++;
            if(l <= r) ans = max(ans,num[i]-num[q[l]-1]);
            while(l <= r && num[i-1] <= num[q[r]-1]) r--;
            q[++r] = i;
        }
        int mx=0;
        fir(i,1,tot){
    
    
            int kk = get_pos(all[i]-k-1)-1,kkk=0;
            if(kk>=1) kkk = get_pos(all[kk]-k-1)-1;
            if(kk >= 1){
    
    
                if(mx < num[kk]-num[max(kkk,0)])
                    mx = num[kk]-num[max(kkk,0)];
            }
            ans = max(ans,mx+num[i]-num[max(kk,0)]);
        }
        cout << ans << endl;
    }    
    
    return 0;
}