2018-2019 ACM-ICPC Pacific Northwest Regional Contest (Div. 1) Solution

A:Exam

Solved.

温暖的签。

#include<bits/stdc++.h>

using namespace std;

const int maxn = 1e3 + 10;

int k;
char str1[maxn], str2[maxn];

int main()
{
    while(~scanf("%d",&k))
    {
        scanf("%s", str1 + 1);
        scanf("%s", str2 + 1);
        int cnt1 = 0, cnt2 = 0;
        int len = strlen(str1 + 1);
        for(int i = 1; i <= len; ++i)
        {
            if(str1[i] == str2[i]) cnt1++;
            else cnt2++;
        }
        int ans = 0;
        if(cnt1 >= k)
        {
            ans = k + cnt2;
        }
        else 
        {
            ans = len - (k - cnt1);
        }
        printf("%d
", ans);
    }
    return 0;
}

B:Coprime Integers

Solved.

枚举gcd反演

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1e7 + 10;

bool check[maxn];
int prime[maxn];
ll mu[maxn];

void Moblus()
{
    memset(check, false, sizeof check);
    mu[1] = 1;
    int tot = 0;
    for(int i = 2; i < maxn; ++i)
    {
        if(!check[i])
        {
            prime[tot++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < tot; ++j)
        {
            if(i * prime[j] > maxn) break;
            check[i * prime[j]] = true;
            if(i % prime[j] == 0)
            {
                mu[i * prime[j]] = 0;
                break;
            }
            else 
            {
                mu[i * prime[j]] = -mu[i];
            }
        }
    }
}

ll sum[maxn];

ll calc(int n, int m)
{
    ll ans = 0;
    if(n > m) swap(n, m);
    for(int i = 1, la = 0; i <= n; i = la + 1)
    {
        la = min(n / (n / i), m / (m / i));
        ans += (ll)(sum[la] - sum[i - 1]) * (n / i) * (m / i);
    }
    return ans;
}    

ll a, b, c, d;

int main()
{
    Moblus();
    for(int i = 1; i < maxn; ++i) sum[i] = sum[i - 1] + mu[i];
    while(~scanf("%lld %lld %lld %lld", &a, &b, &c, &d))
    {
        ll ans = calc(b, d) - calc(a - 1, d) - calc(b, c - 1) + calc(a - 1, c - 1);
        printf("%lld
", ans);
    }
    return 0;
}

C:Contest Setting

Solved.

题意：

有n个题目，每个题目的难度值不同，要选出k个组成一套$Contest$

要求每个题目难度不同，求方案数

思路：

$把难度值相同的题目放在一起作为一种 dp[i][j] 表示选到第i种题目，已经选了k种的方案数$

$然后做01背包即可，注意加上的值为cnt[i] 表示该种题目有多少个$

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll MOD = 998244353;
const int maxn = 1e3 + 10;

int n, k, pos;
ll cnt[maxn];
ll dp[maxn];
map<int, int>mp;

int main()
{
    while(~scanf("%d %d", &n, &k))
    {
        mp.clear();
        pos = 0;
        memset(cnt, 0, sizeof cnt);
        for(int i = 1; i <= n; ++i)
        {
            int num;
            scanf("%d", &num);
            if(mp[num] == 0) mp[num] = ++pos;
            int id = mp[num];
            cnt[id]++;
        }
        memset(dp, 0, sizeof dp);
        dp[0] = 1;
        for(int i = 1; i <= pos; ++i)
        {
            for(int j = k; j >= 1; --j)
            {
                dp[j] = (dp[j] + dp[j - 1] * cnt[i] % MOD) % MOD;
            }
        }
        printf("%lld
", dp[k]);
    }
    return 0;
}

D:Count The Bits

Solved.

题意：

在$[0, 2^b - 1] 中所有k的倍数以二进制形式表示有多少个1$

思路：

$dp[i][j] 表示枚举二进制位数，j 模k的余数， 表示的是前i位中模k的余数为j的数中1的个数$

$cnt[i][j] 表示当前二进制位为第i位， 模k的余数为j的数的个数$

直接转移即可。

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll MOD = (ll)1e9 + 9;
const int maxn = 1e3 + 10;

int k, b;
ll cnt[maxn][maxn];
ll dp[maxn][maxn];

int main()
{
    while(~scanf("%d %d", &k, &b))
    {
        memset(cnt, 0, sizeof cnt);
        memset(dp, 0, sizeof dp);
        cnt[1][0]++;
           cnt[1][1 % k]++;
        dp[1][1 % k]++;
        ll tmp = 1;
        for(int i = 1; i <= b; ++i)
        {
            tmp = (tmp << 1) % k;
            for(int j = 0; j < k; ++j)
            {
                //0
                cnt[i + 1][j] = (cnt[i + 1][j] + cnt[i][j]) % MOD;
                dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
                //1
                ll tmp2 = (tmp + j) % k;
                cnt[i + 1][tmp2] = (cnt[i + 1][tmp2] + cnt[i][j]) % MOD;
                dp[i + 1][tmp2] = ((dp[i + 1][tmp2] + cnt[i][j]) % MOD + dp[i][j]) % MOD;
            }
        }
        printf("%lld
", dp[b][0]);
    }
    return 0;
}

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

#define N 1010
#define ll long long
const ll MOD = (ll)1e9 + 9;
ll dp[N][N], cnt[N][N];
int n, k;

int main()
{
    while (scanf("%d%d", &k, &n) != EOF)
    {
        memset(dp, 0, sizeof dp);
        memset(cnt, 0, sizeof cnt);
        ++dp[1][1 % k];
        ++cnt[1][1 % k];
        ++cnt[1][0];     
        ll tmp = 1 % k;  
        for (int i = 2; i <= n; ++i)
        {
            tmp = tmp * 2 % k; 
            for (int j = 0; j < k; ++j) 
            {
                dp[i][j] = dp[i - 1][j];   
                cnt[i][j] = cnt[i - 1][j]; 
                if (j - tmp >= 0) 
                {
                    dp[i][j] = (dp[i][j] + dp[i - 1][j - tmp] + cnt[i - 1][j - tmp]) % MOD;
                    cnt[i][j] = (cnt[i][j] + cnt[i - 1][j - tmp]) % MOD; 
                }
                else 
                {
                    dp[i][j] = (dp[i][j] + dp[i - 1][k + j - tmp] + cnt[i - 1][k + j - tmp]) % MOD;
                    cnt[i][j] = (cnt[i][j] + cnt[i - 1][k + j - tmp]) % MOD; 
                }
            }    
        }
        printf("%lld
", dp[n][0]);
    }
    return 0;
}

E:Cops And Robbers

Solved.

题意：

在一个$n * m 的矩阵中，有些地方可以建墙，但是不同的墙开销不同，求最小开销把劫匪围住$

思路：

拆点最小割，但是是点权，拆点即可。

#include<bits/stdc++.h>

using namespace std; 

const int maxn = 4e3 + 10;
const int INF = 0x3f3f3f3f;

struct Edge{
    int to, flow, nxt;
    Edge(){}
    Edge(int to, int nxt, int flow):to(to), nxt(nxt), flow(flow){}
}edge[maxn << 2];

int head[maxn], dep[maxn];
int S,T;
int N, n, m, tot;

void Init(int n)
{
    N = n;
    //memset(head, -1, sizeof head);
    for(int i = 0; i < N; ++i) head[i] = -1;
    tot = 0;
}

void addedge(int u, int v, int w, int rw = 0)
{
    edge[tot] = Edge(v, head[u], w); head[u] = tot++;
    edge[tot] = Edge(u, head[v], rw); head[v] = tot++;
}

bool BFS()
{
    //memset(dep, -1, sizeof dep);
    for(int i = 0; i < N; ++i) dep[i] = -1;
    queue<int>q;
    q.push(S);
    dep[S] = 1;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u]; ~i; i = edge[i].nxt)
        {
            if(edge[i].flow && dep[edge[i].to] == -1)
            {
                dep[edge[i].to] = dep[u] + 1;
                q.push(edge[i].to);
            }
        }
    }
    return dep[T] < 0 ? 0 : 1;
}

int DFS(int u, int f)
{
    if(u == T || f == 0) return f;
    int w, used = 0;
    for(int i = head[u]; ~i; i = edge[i].nxt)
    {
        if(edge[i].flow && dep[edge[i].to] == dep[u] + 1)
        {
            w = DFS(edge[i].to, min(f - used, edge[i].flow));
            edge[i].flow -= w;
            edge[i ^ 1].flow += w;
            used += w;
            if(used == f) return f;
        }
    }
    if(!used) dep[u] = -1;
    return used;
}

int Dicnic()
{
    int ans = 0;
    while(BFS())
    {
        int tmp = DFS(S, INF); 
        if(tmp == INF) return -1;
        ans += tmp;
    }
    return ans;
}

int c;
int C[maxn];
char mp[100][100];

int calc(int i, int j)
{
    return i * 31 + j;
}

int main()
{
    while(~scanf("%d %d %d", &m, &n, &c))
    {
        int base = n * 31 + m + 31;
        int x, y;
        Init(4000);
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j <= m; ++j)
            {
                scanf(" %c", &mp[i][j]);
                if(mp[i][j] == 'B') { x = i, y = j; }
            }
        }
        for(int i = 1; i <= c; ++i) scanf("%d", C + i);
        S = 0, T = calc(x, y);
        for(int i = 1; i <= n; ++i)
        {
            addedge(S, calc(i, 1), INF);
            addedge(S, calc(i, m), INF);    
        }
        for(int i = 1; i <= m; ++i)
        {
            addedge(S, calc(1, i), INF);
            addedge(S, calc(n, i), INF);
        }
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j <= m; ++j)
            {
                if(i != n)
                {
                    addedge(calc(i, j) + base, calc(i + 1, j), INF);
                    addedge(calc(i + 1, j) + base, calc(i, j), INF);
                }
                if(j != m)
                {
                    addedge(calc(i, j)+ base, calc(i, j + 1), INF);
                    addedge(calc(i, j + 1) + base, calc(i, j), INF);
                }
            }
        }
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j <= m; ++j)
            {
                if(mp[i][j] >= 'a' && mp[i][j] <= 'z')
                {
                    addedge(calc(i, j), calc(i, j) + base, C[mp[i][j] - 'a' + 1]);
                }
                else 
                {
                    addedge(calc(i, j), calc(i, j) + base, INF);
                }
            }
        }
        int ans = Dicnic();
        printf("%d
", ans);
    }
    return 0;
}

F:Rectangles

Solved.

题意：

二维平面上有一些平行坐标轴的矩形，求有多少区间是被奇数个矩形覆盖。

思路：

扫描线，只是把区间加减换成区间01状态翻转，现场学扫描线可还行..

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

#define ll long long
#define N 200010
#define pii pair <int, int>
int n, mx, my, bx[N], by[N];
ll res;
struct Rec
{
    int x[2], y[2]; 
    void scan()
    {
        for (int i = 0; i < 2; ++i) scanf("%d%d", x + i, y + i);
        if (x[0] > x[1]) swap(x[0], x[1]);
        if (y[0] > y[1]) swap(y[0], y[1]);
        bx[++mx] = x[0];
        bx[++mx] = x[1];
        by[++my] = y[0];
        by[++my] = y[1]; 
    }
}rec[N];
vector <pii> v[N];  

void Hash()
{
    sort(bx + 1, bx + 1 + mx);
    sort(by + 1, by + 1 + my);
    mx = unique(bx + 1, bx + 1 + mx) - bx - 1;
    my = unique(by + 1, by + 1 + my) - by - 1;
    for (int i = 1; i <= n; ++i) 
    {
        for (int j = 0; j < 2; ++j)
        {
            rec[i].x[j] = lower_bound(bx + 1, bx + 1 + mx, rec[i].x[j]) - bx;
            rec[i].y[j] = lower_bound(by + 1, by + 1 + my, rec[i].y[j]) - by; 
        }
    }
}

namespace SEG
{
    struct node
    {
        ll sum, val;
        int lazy; 
        node () {}
        node (ll sum, ll val, int lazy) : sum(sum), val(val), lazy(lazy) {}
        void init() { sum = val = lazy = 0; }
        node operator + (const node &other) const { return node(sum + other.sum, val + other.val, 0); }
        void Xor() { val = sum - val; lazy ^= 1; }   
    }a[N << 2];
    void build(int id, int l, int r)
    {
        a[id] = node(0, 0, 0);
        if (l == r) 
        {
            a[id].sum = by[l + 1] - by[l];
            return;
        }
        int mid = (l + r) >> 1;
        build(id << 1, l, mid);
        build(id << 1 | 1, mid + 1, r);
        a[id] = a[id << 1] + a[id << 1 | 1];
    }
    void pushdown(int id)
    {
        if (!a[id].lazy) return;
        a[id << 1].Xor();
        a[id << 1 | 1].Xor();
        a[id].lazy = 0;
    }
    void update(int id, int l, int r, int ql, int qr)
    {
        if (l >= ql && r <= qr) 
        {
            a[id].Xor();
            return;
        }
        int mid = (l + r) >> 1;
        pushdown(id);
        if (ql <= mid) update(id << 1, l, mid, ql, qr);
        if (qr > mid) update(id << 1 | 1, mid + 1, r, ql, qr);
        a[id] = a[id << 1] + a[id << 1 | 1];
    }
}

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        mx = my = 0; res = 0; 
        for (int i = 1; i <= 2 * n; ++i) v[i].clear();
        for (int i = 1; i <= n; ++i) rec[i].scan(); Hash();    
        for (int i = 1; i <= n; ++i) 
        {
            v[rec[i].x[0]].emplace_back(rec[i].y[0], rec[i].y[1] - 1);
            v[rec[i].x[1]].emplace_back(rec[i].y[0], rec[i].y[1] - 1); 
        }
        n <<= 1;    
        SEG::build(1, 1, n);  
        for (int i = 1; i < mx; ++i)
        {
            for (auto it : v[i]) SEG::update(1, 1, n, it.first, it.second);
            res += (bx[i + 1] - bx[i]) * SEG::a[1].val;    
        }
        printf("%lld
", res); 
    }
    return 0;
}

G:Goat on a Rope

Solved.

签到。

#include<bits/stdc++.h>

using namespace std;

double x, y;
double xa, xb, ya, yb;

double calc(double xa, double ya, double xb, double yb)
{
    return sqrt((xa - xb) * (xa - xb) + (ya - yb) * (ya - yb));
}

int main()
{
    while(~scanf("%lf %lf %lf %lf %lf %lf", &x, &y, &xa, &ya, &xb, &yb))
    {
        double ans = 1e9;
        if(x >= min(xa, xb) && x <= max(xa, xb)) ans = min(ans, min(fabs(y - ya), fabs(y - yb)));
        if(y >= min(ya, yb) && y <= max(ya, yb)) ans = min(ans, min(fabs(x - xa), fabs(x - xb)));
        ans = min(ans, calc(x, y, xa, ya));
        ans = min(ans, calc(x, y, xa, yb));
        ans = min(ans, calc(x, y, xb, ya));
        ans = min(ans, calc(x, y, xb, yb));
        printf("%.3f
", ans);
    }
    return 0;
}

H:Repeating Goldbachs

Solved.

签到。

#include<bits/stdc++.h>

using namespace std;

const int maxn = 1e6 + 10;

bool isprime[maxn];
int prime[maxn];

void Init()
{
    memset(isprime, true, sizeof isprime);
    isprime[0] = isprime[1] = false;
    for(int i = 2; i < maxn; ++i)
    {
        if(isprime[i])
        {
            prime[++prime[0]] = i;
            for(int j = i * 2; j < maxn; j += i)
            {
                isprime[j] = false;
            }
        }
    }
}

int x;

int main()
{
    Init();
    while(~scanf("%d", &x))
    {
        int ans = 0;
        while(x >= 4)
        {
            for(int i = 1; i <= prime[0]; ++i)
            {
                int tmp = prime[i];
                if(isprime[x - tmp])
                {
                    ++ans;
                    x = x - tmp - tmp;
                    break;
                }
            }
        }
        printf("%d
", ans);
    }
    return 0;
}

I:Inversions

Unsolved.

题意：

给出一些序列，一些位置上的数字可以在$[1, k]的范围内随便填，求如果填使得整个序列的逆序对个数最多$

J:Time Limits

Solved.

签到。

#include<bits/stdc++.h>

using namespace std;

int n, s;

int main()
{
    while(~scanf("%d %d", &n, &s))
    {
        int Max = -1;
        for(int i = 1; i <= n; ++i)
        {
            int num;
            scanf("%d", &num);
            Max = max(Max, num);
        }
        Max *= s;
        int ans = Max / 1000;
        if(Max % 1000) ans++;
        printf("%d
", ans);
    }
    return 0;
}

K:Knockout

Unsolved.

题意：

给出一个数字，然后两个骰子的点数，每次可以移掉数字当中某几位加起来的和等于两骰子点数之和

那么就可以移掉这几位，知道最后不能移位置，最后剩下的数就是分数

现在给出一中间局面，求移掉哪些，使得最后的分数期望最大以及最小

L:Liars

Solved.

签到。

#include<bits/stdc++.h>

using namespace std;

const int maxn = 1e3 + 10;

int n;
int cnt[maxn];

int main()
{
    while(~scanf("%d", &n))
    {
        memset(cnt, 0, sizeof cnt);
        for(int i = 1; i <= n; ++i)
        {
            int ai, bi;
            scanf("%d %d", &ai, &bi);
            for(int j = ai; j <= bi; ++j)
            {
                cnt[j]++;
            }
        }
        int ans = -1;
        for(int i = 1; i <= n; ++i)
        {
            if(cnt[i] == i) ans = i;
        }
        printf("%d
", ans);
    }
    return 0;
}

M:Mobilization

Unsolved.

题意：

$有m种军队，每种军队有h_i 属性 和 p_i属性，以及购买一支军队需要c_i的钱，每种军队可以购买无限支$

$现在你有C个单位的钱，求如何购买军队，使得sum h_i cdot sum p_i 最大$