题目来源：https://codeforces.com/contest/1282
脑速不够，差点掉分了，勉强上蓝，算是圣诞礼物了叭~
最可怕的是发生了暑假一样的错误：
两个int型变量相乘然后爆了，希望以后不会再犯了！！！

1282A - Temporarily unavailable

这题可以说考的就是求x轴上两线段的重复的长度，但是我居然debug了18分钟…
我当时AC的代码枚举每一种情况 ~~（我是码农）~~

#include<iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <stack>

#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define r(x) read(x)
#define rrr(x, y, z) read(x);read(y);read(z)
#define mp(x, y) make_pair(x,y)
#define pb(x) push_back(x)
#define all(x) x.begin(),x.end()
#define FOR(i, l, r) for(int i=l;i<=r;i++)
using namespace std;

const int N = 1e5 + 5;
const int M = 1e3 + 5;
const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int SZ = 17;
const double eps = 1e-8;
const double PI = acos(-1);
typedef long long LL;
typedef pair<int, int> pt;

int n, m;
template<class T>
inline void read(T &x) {
    char c;
    x = 1;
    while ((c = getchar()) < '0' || c > '9') if (c == '-') x = -1;
    T res = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
    x *= res;
}
int main() {
    int t;
    r(t);
    while(t--){
        int a,b,c,r;
        rrr(a,b,c); r(r);
        if(a>b) swap(a,b);
        int ans=0;
        int x=c-r;
        int y=c+r;
        if(x<=a&&b<=y){
            cout<<0<<endl;
            continue;
        }
        else if(x<=a&&a<=y){
            cout<<b-y<<endl;
            continue;
        }
        else if(x<=b&&b<=y){
            cout<<x-a<<endl;
            continue;
        }
        if(a<=x&&y<=b){
            cout<<b-a-(y-x)<<endl;
        }
        else{
            cout<<b-a<<endl;
        }
    }
    return 0;
}

优化后简单的代码

#include<iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <stack>

#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define r(x) read(x)
#define rrr(x, y, z) read(x);read(y);read(z)
#define mp(x, y) make_pair(x,y)
#define pb(x) push_back(x)
#define all(x) x.begin(),x.end()
#define FOR(i, l, r) for(int i=l;i<=r;i++)
using namespace std;

const int N = 1e5 + 5;
const int M = 1e3 + 5;
const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int SZ = 17;
const double eps = 1e-8;
const double PI = acos(-1);
typedef long long LL;
typedef pair<int, int> pt;

int n, m;
template<class T>
inline void read(T &x) {
    char c;
    x = 1;
    while ((c = getchar()) < '0' || c > '9') if (c == '-') x = -1;
    T res = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
    x *= res;
}

int main() {
    int t;
    r(t);
    while(t--){
        int a,b,c,r;
        rrr(a,b,c); r(r);
        int x=c-r,y=c+r;
        if(a>b) swap(a,b);
        cout<<b-a-max(0,min(b,y)-max(a,x))<<endl;
    }
    return 0;
}

1282B - K for the Price of One

(Easy & Hard Version)
把价格排序之后不难发现想用最少的钱买最多的肯定是把最靠前的都给买了
我把它当做dp处理了 dp[ i ]表示买前i个最少花多少钱
转移方程dp[i] = min(dp[i], dp[i - k] + f[i])
然后注意 dp[ i ] 不是逐渐增大的 ，我们要找最大的满足的i就ok了故要倒着找

#include<iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <stack>

#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define r(x) read(x)
#define rrr(x, y, z) read(x);read(y);read(z)
#define mp(x, y) make_pair(x,y)
#define pb(x) push_back(x)
#define all(x) x.begin(),x.end()
#define FOR(i, l, r) for(int i=l;i<=r;i++)
using namespace std;

const int N = 1e5 + 5;
const int M = 1e3 + 5;
const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int SZ = 17;
const double eps = 1e-8;
const double PI = acos(-1);
typedef long long LL;
typedef pair<int, int> pt;

int n, m;
int f[N<<2];
LL dp[N<<2];
template<class T>
inline void read(T &x) {
    char c;
    x = 1;
    while ((c = getchar()) < '0' || c > '9') if (c == '-') x = -1;
    T res = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
    x *= res;
}
int main() {
    int t;
    r(t);
    while(t--) {
        r(n);
        r(m);
        int k;
        r(k);
        FOR(i, 1, n) r(f[i]);
        sort(f + 1, f + n + 1);
        int ans = 0;
        dp[0] = 0;
        for (int i = 1; i <= n; i++) {
            dp[i] = dp[i - 1] + f[i];
            if (i >= k) {
                dp[i] = min(dp[i], dp[i - k] + f[i]);
            }
        }
        for(int i=n;i>=0;i--){
            if(m>=dp[i]){
                cout<<i<<endl;
                break;
            }
        }
    }
    return 0;
}

1282C - Petya and Exam

这题考虑起来就是先对ti 排序，每次考虑ti-1的时间能不能容纳前i-1个任务完成，能完成还要看能不能容纳i之后的任务（先考虑简单任务，再考虑复杂任务，这也是贪心的一部分）
~~可怜，我再一次因为爆int而过了pretest没过终测，希望长点记性~~~

#include<iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <stack>

#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define r(x) read(x)
#define rrr(x, y, z) read(x);read(y);read(z)
#define mp(x, y) make_pair(x,y)
#define pb(x) push_back(x)
#define all(x) x.begin(),x.end()
#define FOR(i, l, r) for(int i=l;i<=r;i++)
using namespace std;

const int N = 1e5 + 5;
const int M = 1e3 + 5;
const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int SZ = 17;
const double eps = 1e-8;
const double PI = acos(-1);
typedef long long LL;
typedef pair<int, int> pt;

int n, m;
struct node
{
    int x,y;
}f[N<<2];
bool cmp(node a,node b)
{
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}
template<class T>
inline void read(T &x) {
    char c;
    x = 1;
    while ((c = getchar()) < '0' || c > '9') if (c == '-') x = -1;
    T res = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
    x *= res;
}
int main() {
    int t;
    r(t);
    while(t--) {
        int x,y;
        rrr(n,m,x); r(y);
        int n0=0,n1=0;
        FOR(i,1,n){
            r(f[i].y);
            if(f[i].y) n1++;
            else n0++;
        }
        FOR(i,1,n) r(f[i].x);
        sort(f+1,f+n+1,cmp);
        int ans=0;
        LL need=0;
        f[n+1].x=m+1;
        f[n+1].y=2;
        f[0].x=0;
        f[0].y=2;
        for(int i=0;i<=n;i++){
            if(f[i].y==0){
                need+=x;
                n0--;
            }
            else if(f[i].y==1){
                need+=y;
                n1--;
            }
            if(need>m) break;
            else if(need<=f[i+1].x-1) {
                int now=i;
                LL res=f[i+1].x-1-need;
                if(1ll*x*n0>res){
                    now+=res/x;
                }
                else{
                    now+=n0;
                    res-=1ll*n0*x;
                    if(1ll*y*n1>res){
                        now+=res/y;
                    }
                    else{
                        now+=n1;
                    }
                }
                ans=max(ans,now);
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

1282D - Enchanted Artifact

这题是一道特殊的人机互动题，~~什么是互动题自行查找~~
我也是比赛前一小时才知道什么是互动题的，可惜前面卡太久了，这些题没时间做了 555 算了放一个介绍链接吧
怎么在n+2次查找中找到答案呢？这个n+2很有深意哦~
思路如下：
第一次问 300个a 得 q1
第二次问 300个b 得 q2
想一想 N-q1是不是最终的a的个数 N-q2是不是最终的b的个数 总长度n=a+b
然后假设最终答案为 1个b加n-1个a ，此时如果真正的答案的第一个字符确实是b 那会返回b-1 否则返回b+1 通过这种区别来往后判断这个位置是b还是a
值得注意的是你写出的代码可能最多循环了n+3次 还需要特殊处理见代码不细说

#include<iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <stack>

#define ls k<<1,l,mid
#define rs k<<1|1,mid+1,r
#define r(x) read(x)
#define rrr(x, y, z) read(x);read(y);read(z)
#define mp(x, y) make_pair(x,y)
#define pb(x) push_back(x)
#define all(x) x.begin(),x.end()
#define FOR(i, l, r) for(int i=l;i<=r;i++)
using namespace std;

const int N = 1e5 + 5;
const int M = 1e3 + 5;
const int mod = 1e9 + 7;
const int INF = 0x7fffffff;
const int SZ = 17;
const double eps = 1e-8;
const double PI = acos(-1);
typedef long long LL;
typedef pair<int, int> pt;

int n, m;
template<class T>
inline void read(T &x) {
    char c;
    x = 1;
    while ((c = getchar()) < '0' || c > '9') if (c == '-') x = -1;
    T res = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
    x *= res;
}
int hhh(string s)
{
    cout<<s<<endl;
    int op;
    r(op);
    if(op==0) exit(0);
    return op;
}
int main() {
    int LEN=300;
    int rest=hhh(string(LEN,'a'));
    n=2*LEN-rest-hhh(string(LEN,'b'));
    int a=LEN-rest,b=n-a;
    string ans(n,'a');
    int tmp=b;
    for(int i=0;i<n-1;i++){
        ans[i]='b';
        if(hhh(ans)>tmp){
            ans[i]='a';
        }
        else tmp--;
    }
    if(tmp) ans[n-1]='b';
    hhh(ans);
    return 0;
}

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Codeforces Round #610 (Div. 2)【A - D】

1282A - Temporarily unavailable

1282B - K for the Price of One

1282C - Petya and Exam

1282D - Enchanted Artifact

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