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 = 1; 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 0;
} 

 

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 << 11;
    x ^= x >> 4;
    x ^= x << 5;
    x ^= x >> 14;
    ui p = x ^ (y ^ z);
    x = y;
    y = z;
    z = p;
    return z;
}

ll zx[400005];
int lz[400005];

void pushup(int rt)
{
    zx[rt] = min(zx[rt << 1], zx[rt << 1 | 1]);
}

void pushdown(int rt)
{
    if(lz[rt])
    {
        lz[rt << 1] = max(lz[rt << 1], lz[rt]);
        lz[rt << 1 | 1] = max(lz[rt << 1 | 1], lz[rt]);
        zx[rt << 1] = max(zx[rt << 1], (ll)lz[rt << 1]);
        zx[rt << 1 | 1] = max(zx[rt << 1 | 1], (ll)lz[rt << 1 | 1]);
        lz[rt] = 0;
    }
}

void build(int l, int r, int rt)
{
    if(l == r)
    {
        zx[rt] = lz[rt] = 0;
        return;
    }
    zx[rt] = lz[rt] = 0;
    
    int m = l + r >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
}

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 >> 1;
    if(ql <= m) update(ql, qr, v, l, m, rt << 1);
    if(qr > m) update(ql, qr, v, m + 1, r, rt << 1 | 1);
    pushup(rt);
}

ll query(int l, int r, int rt)
{
    if(l == r) return 1LL * (ll)l * zx[rt]; 
    pushdown(rt);
    
    ll res = 0;
    int m = l + r >> 1;
    res ^= query(l, m, rt << 1);
    res ^= query(m + 1, r, rt << 1 | 1);
    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(1, n, 1);
        for(int i = 1; i <= m; i++)
        {
            ui f1 = sui(), f2 = sui(), f3 = sui();
            l = min(f1 % n + 1, f2 % n + 1);
            r = max(f1 % n + 1, f2 % n + 1);
            v = f3 % (1 << 30);
            update(l, r, v, 1, n, 1);
        }
        printf("%lld\n", query(1, n, 1));
    }
    return 0;
}

 

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[100005];
int a[100005];
int b[15];
int dp[100005][15];
int l[100005][15];
int r[100005][15];
int n, cnt, nl, nr;
int ans, ansl, ansr;

void solve(int len)
{
    for(int i = 1; i <= len; i++) dp[0][i] = 0;    
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= len; j++)
        {
            dp[i][j] = dp[i - 1][j];
            l[i][j] = l[i - 1][j];
            r[i][j] = r[i - 1][j];
            if(a[i] == b[j])
            {
                dp[i][j]++;
                if(j == nl && l[i][j] == 0) l[i][j] = i;
                if(j == nr) r[i][j] = i;
            }
            
            if(dp[i][j - 1] > dp[i][j])
            {
                dp[i][j] = dp[i][j - 1];
                l[i][j] = l[i][j - 1];
                r[i][j] = r[i][j - 1];
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        scanf("%s", s);
        int zx = 9, zd = 0;
        for(int i = 1; i <= 10; i++) b[i] = i - 1;
        for(int i = 0; i < n; i++)
        {
            a[i + 1] = s[i] - '0';
            zx = min(zx, a[i + 1]);
            zd = max(zd, a[i + 1]);
        }
        
        solve(10);
        ansl = 1;
        ansr = 1;
        ans = dp[n][10];
        for(int i = zx; i <= zd; i++)
        {
            for(int j = zx; j < i; j++)
            {    
                cnt = 0;
                for(int a = 0; a <= j; a++) b[++cnt] = a;
                nl = cnt + 1;
                for(int a = i; a >= j; a--) b[++cnt] = a;
                nr = cnt;
                for(int a = i; a < 10; 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 0;
}