HDU多校Round 5

Solved:3

rank:71

E. Everything Has Changed

#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int m; double R;
        scanf("%d%lf", &m, &R);

        double ans = 2.0 * PI * R;
        for(int i = ; i <= m; i++)
        {
            double x, y, r;
            scanf("%lf%lf%lf", &x, &y, &r);

            double dis = sqrt(x * x + y * y);
            if(dis + r < R || dis - r >= R) continue;

            double c1 = acos((r * r + dis * dis - R * R) / (2.0 * dis * r));
            ans += c1 * 2.0 * r;
            double c2 = acos((R * R + dis * dis - r * r) / (2.0 * dis * R));
            ans -= c2 * 2.0 * R;
        }
        printf("%.20lf\n", ans);
    }
    return ;
} 

G. Glad You Came

#include <bits/stdc++.h>
using namespace std;
typedef unsigned int ui;
typedef long long ll;

ui x, y, z;
ui sui()
{
    x ^= x << ;
    x ^= x >> ;
    x ^= x << ;
    x ^= x >> ;
    ui p = x ^ (y ^ z);
    x = y;
    y = z;
    z = p;
    return z;
}

ll zx[];
int lz[];

void pushup(int rt)
{
    zx[rt] = min(zx[rt << ], zx[rt <<  | ]);
}

void pushdown(int rt)
{
    if(lz[rt])
    {
        lz[rt << ] = max(lz[rt << ], lz[rt]);
        lz[rt <<  | ] = max(lz[rt <<  | ], lz[rt]);
        zx[rt << ] = max(zx[rt << ], (ll)lz[rt << ]);
        zx[rt <<  | ] = max(zx[rt <<  | ], (ll)lz[rt <<  | ]);
        lz[rt] = ;
    }
}

void build(int l, int r, int rt)
{
    if(l == r)
    {
        zx[rt] = lz[rt] = ;
        return;
    }
    zx[rt] = lz[rt] = ;

    int m = l + r >> ;
    build(l, m, rt << );
    build(m + , r, rt <<  | );
}

void update(int ql, int qr, int v, int l, int r, int rt)
{
    if(zx[rt] >= v) return;
    if(ql <= l && qr >= r)
    {
        lz[rt] = max(lz[rt], v);
        zx[rt] = max(zx[rt], (ll)v);
        return;
    }

    pushdown(rt);
    int m = l + r >> ;
    if(ql <= m) update(ql, qr, v, l, m, rt << );
    if(qr > m) update(ql, qr, v, m + , r, rt <<  | );
    pushup(rt);
}

ll query(int l, int r, int rt)
{
    if(l == r) return 1LL * (ll)l * zx[rt];
    pushdown(rt);

    ll res = ;
    int m = l + r >> ;
    res ^= query(l, m, rt << );
    res ^= query(m + , r, rt <<  | );
    return res;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, m;
        cin>>n>>m>>x>>y>>z;
        int l, r, v;

        build(, n, );
        for(int i = ; i <= m; i++)
        {
            ui f1 = sui(), f2 = sui(), f3 = sui();
            l = min(f1 % n + , f2 % n + );
            r = max(f1 % n + , f2 % n + );
            v = f3 % ( << );
            update(l, r, v, , n, );
        }
        printf("%lld\n", query(, n, ));
    }
    return ;
}

H. Hills And Valleys

构造一个0-9的数组 每次枚举翻转哪两个数就一共45次

然后把两个数组跑一个lcs 枚举翻转的那个数组可以多次匹配

#include <bits/stdc++.h>
using namespace std;

char s[];
int a[];
int b[];
int dp[][];
int l[][];
int r[][];
int n, cnt, nl, nr;
int ans, ansl, ansr;

void solve(int len)
{
    for(int i = ; i <= len; i++) dp[][i] = ;
    for(int i = ; i <= n; i++)
    {
        for(int j = ; j <= len; j++)
        {
            dp[i][j] = dp[i - ][j];
            l[i][j] = l[i - ][j];
            r[i][j] = r[i - ][j];
            if(a[i] == b[j])
            {
                dp[i][j]++;
                if(j == nl && l[i][j] == ) l[i][j] = i;
                if(j == nr) r[i][j] = i;
            }

            if(dp[i][j - ] > dp[i][j])
            {
                dp[i][j] = dp[i][j - ];
                l[i][j] = l[i][j - ];
                r[i][j] = r[i][j - ];
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        scanf("%s", s);
        int zx = , zd = ;
        for(int i = ; i <= ; i++) b[i] = i - ;
        for(int i = ; i < n; i++)
        {
            a[i + ] = s[i] - '';
            zx = min(zx, a[i + ]);
            zd = max(zd, a[i + ]);
        }

        solve();
        ansl = ;
        ansr = ;
        ans = dp[n][];
        for(int i = zx; i <= zd; i++)
        {
            for(int j = zx; j < i; j++)
            {
                cnt = ;
                for(int a = ; a <= j; a++) b[++cnt] = a;
                nl = cnt + ;
                for(int a = i; a >= j; a--) b[++cnt] = a;
                nr = cnt;
                for(int a = i; a < ; a++) b[++cnt] = a;
                solve(cnt);

                if(dp[n][cnt] > ans && l[n][cnt] && r[n][cnt])
                {
                    ans = dp[n][cnt];
                    ansl = l[n][cnt];
                    ansr = r[n][cnt];
                }
            }
        }
        printf("%d %d %d\n", ans, ansl, ansr);
    }
    return ;
}