长 乐 集 训 D a y 3 – 小插曲

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关于阳辰楠抄袭标程得到 3/4 AK的通报批评

近日，阳辰楠同学场场比赛都抄袭标程，以此获得高分。这件事情遭到了许多同学的不满。还屡教不改，不承认自己抄标程的事情。我们在此通报批评阳辰楠同学。对他的行为提出强烈的谴责！希望其他同学可以引以为戒，诚信考试，不能抄标程。做一名诚信的OIer！

附录：感谢yjy同学的以身试法，让yy发现了ycn一直抄标程的肮脏行径。

详情可见

证据

「Noip2023国庆集训A组Day2」叉叉计算

//YCN的代码
#include 
using namespace std;
const int N = 100010;
int len, ans;
string s;
map vis;
map val;
stack s1, st2;
int main() {
    freopen("cross.in", "r", stdin);
    freopen("cross.out", "w", stdout);
    cin >> s;
    len = s.size();
    for (int i = 0; i < len; i++) {
        if (!vis[s[i]]) val[s[i]] = 0;
        if (s1.empty()) s1.push(s[i]), vis[s[i]] = 1;
        else {
            if (!s1.empty() && s1.top() == s[i]) {
                vis[s[i]] = 0;
                ans += val[s[i]];
                s1.pop();
                continue;
            } else if (s1.top() != s[i] && !vis[s[i]]) {
                vis[s[i]] = 1;
                s1.push(s[i]);
                continue;
            } else if (vis[s[i]] && !s1.empty()) {
                while (s1.top() != s[i]) {
                    int u = s1.top();
                    s1.pop();
                    st2.push(u);
                    val[u]++;
                }
                s1.pop();
                vis[s[i]] = 0;
                ans += val[s[i]];
                val[s[i]] = 0;
                while (!s1.empty()) st2.push(s1.top()), s1.pop();
                while (!st2.empty()) {
                    int u = st2.top();
                    st2.pop();
                    s1.push(u);
                }
            }
        }
    }
    cout << ans;
    return 0;
}

//标准程序 https://blog.csdn.net/weixin_30793643/article/details/97311282
#include
#include
#include
#include
#include

#define N 100007

using namespace std;
int n,ans;
char s[N];
mapvis;
mapval;
stackst1;
stackst2;

int main()
{
    freopen("cross.in","r",stdin);
    freopen("cross.out","w",stdout);
    scanf("%s",s);n=strlen(s);
    for(int i=0;i<n;i++)
    {
        if(!vis[s[i]]) val[s[i]]=0;
        if(st1.empty()) st1.push(s[i]),vis[s[i]]=1;
        else
        {
            if(!st1.empty() && st1.top()==s[i])
            {
                vis[s[i]]=0;ans+=val[s[i]];st1.pop();continue;
            }
            else if(st1.top()!=s[i] && !vis[s[i]])
            {
                vis[s[i]]=1;st1.push(s[i]);continue;
            }
            else if(vis[s[i]] && !st1.empty())
            {
                while(st1.top()!=s[i])
                {
                    int u=st1.top();st1.pop();
                    st2.push(u);val[u]++;
                }
                st1.pop();vis[s[i]]=0;ans+=val[s[i]];val[s[i]]=0;
                while(!st1.empty()) st2.push(st1.top()),st1.pop();
                while(!st2.empty())
                {
                    int u=st2.top();st2.pop();
                    st1.push(u);
                }
            }
        }
    }
    printf("%d\n",ans);
    fclose(stdin);fclose(stdout);
    return 0;
}

证据2

「Noip2023国庆集训A组Day3」古老谜题

//ycn代码
#include 
using namespace std;
const int N = 1e6 + 10;
typedef long long ll;
ll op, n, a[N], num[2], ne[N], sum[N], ans, cnt;
char c[N];
int main() {
    freopen("puzzle.in", "r", stdin);
    freopen("puzzle.out", "w", stdout);
    cin >> op >> n;
    for (int i = 1; i > c[i];
        a[i] = c[i] - '0';
        sum[i] = (a[i] + sum[i - 1]) % 2;
        num[sum[i]]++;
    }
    cnt = n + 1;
    for (int i = n; i >= 1; i--) {
        ne[i] = cnt;
        if (a[i] == 1) cnt = i;
    }
    for (int i = 1; i  1) num[sum[i - 1]]--;
        ans += num[sum[i - 1] ^ 1];
        if (a[i] == 0 && ne[i] <= n) ans--;
        else if (a[i] == 1) ans -= ne[i] - i;
    }
    cout << ans;
    return 0;
}

//标准程序 https://www.cnblogs.com/stoorz/p/13803822.html
#include 
using namespace std;
typedef long long ll;

const int N=1000010;
int type,n,cnt[2],a[N],nxt[N],sum[N];
ll ans;

int main()
{
    freopen("puzzle.in","r",stdin);
    freopen("puzzle.out","w",stdout);
    scanf("%d%d",&type,&n);
    for (int i=1;i=1;i--)
    {
        nxt[i]=last;
        if (a[i]==1) last=i;
    }
    for (int i=1;i1) cnt[sum[i-1]]--;
        ans+=cnt[sum[i-1]^1];
        if (!a[i] && nxt[i]<=n) ans--;
        if (a[i]) ans-=(nxt[i]-i);
    }
    printf("%lld",ans);
    return 0;
}

证据3

「Noip2023国庆集训A组Day3」礼物购买
这一条在讲题时被通报了

//ycn的代码
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long

using namespace std;
inline LL read() {
    LL d = 0, f = 1;
    char s = getchar();
    while (s  '9') {
        if (s == '-')
            f = -1;
        s = getchar();
    }
    while (s >= '0' && s  y.v; }
int main() {
    freopen("present.in", "r", stdin);
    freopen("present.out", "w", stdout);
    LL n = read(), m = read();
    for (LL i = 1; i <= n; i++) {
        a[i].v = read();
        a[i].x = read();
        b[i] = a[i].v;
    }
    sort(b + 1, b + 1 + n);
    sort(a + 1, a + 1 + n, cmp);
    for (LL i = 1; i <= n; i++) {
        sum[i] = sum[i - 1] + a[i].x * a[i].v;
        cnt[i] = cnt[i - 1] + a[i].x;
    }
    sum[n + 1] = b[n + 1] = a[n + 1].v = 1e18;
    while (m--) {
        LL w = read();
        ans = 0;
        for (LL i = 1; i <= n;) {
            LL r = upper_bound(sum + 1, sum + 2 + n, sum[i - 1] + w) - sum;
            w -= sum[r - 1] - sum[i - 1];
            ans += cnt[r - 1] - cnt[i - 1];
            ans += w / a[r].v;
            w %= a[r].v;
            i = n - (upper_bound(b + 1, b + n - r + 1, w) - b - 1) + 1;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

笑着写检讨书的人，最为可恶。

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