CFhM xjb training 题解

第二期第四次每周训练题解

Authored by CFhM_R, 2011 - 2017, All Rights Reserved.

[[e-mail]](my mail)

A-Moon Safari (medium)-数论

• dp[n][r]=∑i=1naiir=∑i=0n−1ai+1(i+1)r=a∑i=0n−1ai∑k=0rCkrik=a∑i=1n−1ai∑k=0rCkrik=a∑k=0rCkr∑i=1n−1aiik=a∑k=0rCkrdp[n−1][k]

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef long long LL;const LL mod = 1e9 + 7;const int maxn = 446;LL C[maxn][maxn];struct Matrix{    LL m[maxn][maxn];    Matrix(){memset(m, 0, sizeof(m));}    void E(){memset(m, 0, sizeof(m)); for(int i = 0; i < maxn; i++) m[i][i] = 1;}};Matrix M_mul(Matrix a, Matrix b, int r){    Matrix ret;    for(int i = 0; i <= r; i++)        for(int k = 0; k <= r; k++)            if(a.m[i][k])                for(int j = 0; j <= r; j++)                    if(b.m[k][j])                        ret.m[i][j] = ( ret.m[i][j] + a.m[i][k] * b.m[k][j]) % mod;    return ret;}Matrix M_qpow(Matrix P, LL n, int r){    Matrix ret;    ret.E();    while(n)    {        if(n & 1LL) ret = M_mul(ret, P, r);        n >>= 1LL;        P = M_mul(P, P, r);    }    return ret;}int main(void){    for(int i = 0; i < maxn; i++) C[i][0] = C[i][i] = 1;    for(int i = 2; i < maxn; i++)        for(int j = 1; j <= i; j++)            C[i][j] = (C[i-1][j-1] + C[i-1][j]) % mod;    int T;    scanf("%d", &T);    while(T--)    {        int N, a, r;        scanf("%d %d %d", &N, &a, &r);        Matrix M;        for(int i = 0; i <= r; i++)            for(int j = 0; j <= i; j++)                M.m[i][j] = C[i][j] * a % mod;        M.m[r+1][r] = M.m[r+1][r+1] = 1;        M = M_qpow(M, N, r + 1);        LL ans = 0LL;        for(int i = 0; i <= r; i++) ans = (ans + M.m[r+1][i] * a) % mod;        printf("%lld\n", ans);    }    return 0;}

---

B-Number Busters-推公式

• 当b<x时，a和c都会减小，c−a不会改变，所以只有在b≥x的时候，才会使a和c之间的距离缩短，所以我们知道b≥x这个状态共出现c−a秒。
• 再来看b<x的时间，假设有k秒，且易知使c=a这个临界点出现的状态必然是b≥x，此时（更新之后）b的值为b−x∗(c−a)+k∗(w−x)，那么这个值还原回去（加上x）必须大于等于x。
• 由此得出方程b−x∗(c−a)+k∗(w−x)+x≥x
• 解出k，再加上b≥x的时间c−a，就是最终的答案

int main() {    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);        long _begin_time = clock();    #endif    ll a, b, w, x, c;    scan(a, b, w); scan(x, c);    ll ans = 0LL;    if(c > a) ans = ceil(1.0 * ((c - a) * x - b) / (w - x)) + c - a;    dbg(ans);    #ifndef ONLINE_JUDGE        long _end_time = clock();        printf("time = %ld ms\n", _end_time - _begin_time);    #endif    return 0;}

---

C-ZYB loves Xor I-分治

• 这里的lowbit(x)的定义和我们以前知道的那个一样
• 网上的题解都是字典树，然而学艺不精
• 刘庆晖老师教导的分治的思维，让我看到了希望
• 对于一个数组a，我们考虑它的最低位是否为1，可以分为两个集合，这两个集合中任意两个数做异或的结果的lowbit，都是1。
• 对于第一个集合，由于最低位相同（都是1），那么考虑第二位（次低位），按照上面的描述再次分为两个集合，那么这两个集合中任意两个数的异或的lowbit都是2。
• 对于第二个集合同理。
• 之后对于再次划分的集合我们发现依然可以这样分治下去。
• wow~
• 最后我们构造的就是一棵2n−1个节点的解答树
• 复杂度是O(nlogn∗T)

void gao(vi &vec, int pos, ll &ans) {    vi vec1, vec2;    for(auto v : vec) {        if(v & (1 << pos)) vec1.pb(v);        else vec2.pb(v);    }    ans += (ll)(1LL << pos) * vec1.size() * vec2.size() % MOD;    if(pos >= 28) return;    if(vec1.size() > 0) gao(vec1, pos + 1, ans);   //剪枝很重要    if(vec2.size() > 0) gao(vec2, pos + 1, ans);}int main() {    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);        long _begin_time = clock();    #endif    int T; scan(T);    rep(i, 0, T)    {        vi vec;        int n, x; scan(n);        while(n--) {            scan(x);            vec.pb(x);        }        ll ans = 0LL;        gao(vec, 0, ans);        ans = ans * 2LL % MOD;        printf("Case #%d: %lld\n", i + 1, ans);    }    #ifndef ONLINE_JUDGE        long _end_time = clock();        printf("time = %ld ms\n", _end_time - _begin_time);    #endif    return 0;}

D-Wavy numbers-Q神代码

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#include<unordered_map>using namespace std;typedef long long ll;const int MAXQ=10000000;const int MAXL=14;int vis[MAXQ],num[MAXL];int change(int t){    int now=0;    while(t)    {        num[now++]=t%10;        t/=10;    }    return now;}unordered_map<ll,int>mp[2][10];void solve_right(ll n,ll &k){    for(int i=1;i<MAXQ;i++)    {        int t=change(i),ok=1;        if(t==2)ok&=(num[0]!=num[1]);        for(int j=1;j<t-1;j++)            ok&=((num[j]>num[j-1] && num[j]>num[j+1])               ||(num[j]<num[j-1] && num[j]<num[j+1]));        if(!ok)continue;        vis[i]=1;        k-=(i%n==0);        if(k==0)        {            printf("%d",i);            exit(0);        }        if(t==6 && num[t-1]>num[t-2])mp[0][0][i%n]++;        else if(t==7)mp[num[t-1]>num[t-2]][num[t-1]][i%n]++;    }}void get_res(ll n,ll &k,ll Mod,int lef){    int len=7;    while(lef)    {        num[len++]=lef%10;        lef/=10;    }    for(int i=1;i<MAXQ;i++)    {        if(!vis[i])continue;        int t=change(i),ok=1;        while(t<7)num[t++]=0;        t=len;        for(int j=1;j<t-1;j++)            ok&=((num[j]>num[j-1] && num[j]>num[j+1])               ||(num[j]<num[j-1] && num[j]<num[j+1]));        if(!ok)continue;        k-=(i%n==Mod);        if(k==0)        {            printf("%07d",i);            exit(0);        }    }}void solve_left(ll n,ll &k){    for(int i=1;i<MAXQ;i++)    {        if(!vis[i])continue;        int t=change(i),ok=1;        if(t==2)ok&=(num[0]!=num[1]);        for(int j=1;j<t-1;j++)            ok&=((num[j]>num[j-1] && num[j]>num[j+1])               ||(num[j]<num[j-1] && num[j]<num[j+1]));        if(!ok)continue;        ll m=(n-1LL*i*MAXQ%n)%n,cnt=0;        if(t==1)        {            for(int j=0;j<num[0];j++)                if(mp[0][j].find(m)!=mp[0][j].end())                    cnt+=mp[0][j][m];            for(int j=num[0]+1;j<10;j++)                if(mp[1][j].find(m)!=mp[1][j].end())                    cnt+=mp[1][j][m];        }        else        {            int go=(num[1]>num[0]);            if(go==0)for(int j=0;j<num[0];j++)                if(mp[0][j].find(m)!=mp[0][j].end())                    cnt+=mp[0][j][m];            if(go==1)for(int j=num[0]+1;j<10;j++)                if(mp[1][j].find(m)!=mp[1][j].end())                    cnt+=mp[1][j][m];        }        if(k<=cnt)        {            printf("%d",i);            get_res(n,k,m,i);        }        else k-=cnt;    }}int main(){    ll n,k;    scanf("%lld%lld",&n,&k);    solve_right(n,k);    solve_left(n,k);    return 0*printf("-1");}

---

E-Bear and Floodlight-计算几何

• 只有20盏灯，所以可以考虑用状压dp，dp[i]表示i代表的状态（1表示亮）能走的最远距离
• 转移就是枚举剩下的未亮的灯，点亮后能走的距离。
• 对于状态dp[i]点亮灯j之后的状态分两种情况：
  
  • 由dp[i]的终点开始向r照亮θj的范围，那么有dp[i|(1<<j)]=(arctan((dp[i]−xj)/yj)+θj)∗yj+xj−l
  • 若该角度超过r，则取r−l

double l, r;struct point {    double x, y, ang;    point() {}    point(double _x, double _y, double _ang) {        x = _x - l;        y = fabs(_y);        ang = _ang * pi / 180.0;    }}p[maxn];double dp[maxn];int main() {    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);        long _begin_time = clock();    #endif    int n; scan(n); scan(l, r);    r -= l;    rep(i, 0, n) {        double x, y, ang;        scan(x, y, ang);        p[i] = point(x, y, ang);    }    rep(i, 0, (1 << n)) {        rep(j, 0, n) {            if((i & (1 << j)) == 0) {                double tmp = atan((r - p[j].x) / p[j].y);                tmp = min(tmp, atan((dp[i] - p[j].x) / p[j].y) + p[j].ang);                dp[i | (1 << j)] = max(dp[i | (1 << j)], p[j].x + p[j].y * tan(tmp));            }        }    }    printf("%.9f\n", dp[(1 << n) - 1]);    #ifndef ONLINE_JUDGE        long _end_time = clock();        printf("time = %ld ms\n", _end_time - _begin_time);    #endif    return 0;}

F-Subway Innovation-前缀和优化+数学推导

• 如果这n个点是按照x坐标排好序的，那么我们很容易知道，两两距离之和最短的k个点一定是连续的（反证一下即可）
• 首先用前缀和处理一下x坐标的和，即sum[i]表示前i个点的x坐标之和
• 令f(i)表示从i开始的k个点两两之间的距离之和，我们先求出f(0)
• 添加辅助函数h(i)表示前i个点两两间的距离之和，有h(i)=h(i−1)+xi∗i−sum[i−1](i从0开始)
• 我们令f(0)=h(k−1)
• 接下来推导f的转移关系，可以看出f(i)到f(i−1)，多出了xi+k−1到xi...xi+k−2的距离，减少了xi−1到xi...xi+k−2的距离
• f(i)=f(i−1)−(sum[i+k−2]−sum[i−1]−xi−1∗(k−1))+xi+k−2∗k−sum[i+k−1]+sum[i−1]

struct point {    int id;    ll x;    bool operator<(const point &b) const {        return x < b.x;    }}p[maxn];ll sum[maxn], dp[maxn];int main() {    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);        long _begin_time = clock();    #endif    int n, k; scan(n);    rep(i, 0, n) {        scan(p[i].x);        p[i].id = i;    }    sort(p, p + n);    sum[0] = p[0].x;    rep(i, 1, n) sum[i] = sum[i - 1] + p[i].x;    scan(k);    rep(i, 1, k) dp[i] = dp[i - 1] + p[i].x * i - sum[i - 1];    dp[0] = dp[k - 1];    ll ans = dp[0], l = 0;    rep(i, 1, n - k + 1) {        dp[i] = dp[i - 1] - (sum[i + k - 2] - sum[i - 1] - p[i - 1].x * (k - 1)) + (p[i + k - 1].x * (k - 1) - (sum[i + k - 2] - sum[i - 1]));        if(ans > dp[i]) {            ans = dp[i];            l = i;        }    }    rep(i, l, l + k) printf("%d%c", p[i].id + 1, i == l + k - 1 ? '\n' : ' ');    #ifndef ONLINE_JUDGE        long _end_time = clock();        printf("time = %ld ms\n", _end_time - _begin_time);    #endif    return 0;}

---

G-Dexterina’s Lab-矩阵优化+概率dp

• dp[i][j]=∑k=0127dp[i−1][k]∗p[j∧k]
• ⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜dp[i][0]dp[i][1]...dp[i][127]⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜p[0][0]p[1][0]...p[127][0]p[0][1]p[1][1]...p[127][1]..................p[0][127]p[1][127]...p[127][127]⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜dp[i−1][0]dp[i−1][1]...dp[i−1][127]⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟

int n, K;struct Matrix{    double x[128][128];}A,B,ans;Matrix operator * (const Matrix &k1, const Matrix &k2) {    mem(B.x, 0);    rep(i, 0, n + 1)        rep(j, 0, n + 1)            rep(k, 0, n + 1)                B.x[i][j] += k1.x[i][k] * k2.x[k][j];    return B;}int main() {    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);        long _begin_time = clock();    #endif    scan(K, n);    rep(i, 0, n + 1) {        double k1; scan(k1);        rep(j, 0, 128) A.x[j][i^j]=k1;    }    n = 127;    ans=A; K--;    while (K) {        if (K & 1) ans = ans * A; A = A * A; K >>= 1;    }    printf("%.11lf\n",1 - ans.x[0][0]);    #ifndef ONLINE_JUDGE        long _end_time = clock();        printf("time = %ld ms\n", _end_time - _begin_time);    #endif    return 0;}

---

H-Count Good Substrings-机智

• 根据描述最后的串一定是abababa这个样子
• 若是回文串那么首尾字母一定相同
• 偶数长度的回文串的来源
  
  • 一个奇数位置的a和一个偶数位置的a之间
  • 一个奇数位置的b和一个偶数位置的b之间
• 奇数长度的回文串来源
  
  • 任意位置的奇偶性相同的两个a或两个b之间的串
  • 单一字符

ll gao(ll n) { return n * (n - 1) / 2LL; }char s[maxn];int main() {    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);        long _begin_time = clock();    #endif    scan(s);    int n = strlen(s);    ll oa = 0LL, ea = 0LL, ob = 0LL, eb = 0LL;    rep(i, 0, n) {        if(i & 1) {            if(s[i] == 'a') oa++;            else ob++;        } else {            if(s[i] == 'a') ea++;            else eb++;        }    }    dbg(oa * ea + ob * eb);    dbg(gao(oa) + gao(ob) + gao(ea) + gao(eb) + n);    #ifndef ONLINE_JUDGE        long _end_time = clock();        printf("time = %ld ms\n", _end_time - _begin_time);    #endif    return 0;}

---

I-Divisors-模拟

• 先O(x√)预处理出x的所有因子并且排序，再根据题目描述dfs模拟一下就行
• 注意对因子去重什么的，免得平方根因子不好搞

int cnt = 0, mnt = 0;ll fac[maxn];void gao(ll x, ll k) {    if(cnt > 99999) return;    if(x == 1LL || k == 0LL) { dbg(x); cnt++; return; }    rep(i, 0, mnt) {        if(x < fac[i]) break;        if(x % fac[i] == 0) gao(fac[i], k - 1);    }}int main() {    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);        long _begin_time = clock();    #endif    ll x, k; scan(x, k);    for(ll i = 1LL; i * i <= x; i++) {        if(x % i == 0) {            fac[mnt++] = i;            fac[mnt++] = x / i;        }     }    mnt = unique(fac, fac + mnt) - fac;    sort(fac, fac + mnt);    gao(x, k);    #ifndef ONLINE_JUDGE        long _end_time = clock();        printf("time = %ld ms\n", _end_time - _begin_time);    #endif    return 0;}

---

J-Painting Fence-分治

• 一个问题的定义，就是找到这个区间中的最短板，把这个长度以下部分横着刷，剩下的以他为中心分为两个区间，每个区间又是相同的子问题

int a[maxn];int gao(int l, int r, int minn) {    if(l > r) return 0;    if(l == r) return a[l] > minn;    int m = min_element(a + l, a + r + 1) - a;    return min(r - l + 1, gao(l, m - 1, a[m]) + gao(m + 1, r, a[m]) + a[m] - minn);}int main() {    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);        freopen("out.txt", "w", stdout);        long _begin_time = clock();    #endif    int n; scan(n);    rep(i, 1, n + 1) scan(a[i]);    dbg(gao(1, n, 0));    #ifndef ONLINE_JUDGE        long _end_time = clock();        printf("time = %ld ms\n", _end_time - _begin_time);    #endif    return 0;}

---

头文件

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <bits/stdc++.h>using namespace std;#define fi first#define se second#define MP(A, B) make_pair(A, B)#define pb push_back#define gcd __gcd#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define rep(i, a, b) for(int i = a; i < b; i++)#define per(i, a, b) for(int i = a; i > b; i--)typedef long long ll;typedef unsigned long long ulls;typedef unsigned int uint;typedef pair<int, int> pii;typedef vector<int> vi;typedef vector<pii> vii;typedef map<int, int> mii;typedef map<string, int> msi;typedef map<pii, int> mpi;const int INF = 0x3f3f3f3f;const ll LINF = 0x3f3f3f3f3f3f3f3fLL;const ll MOD = 998244353;const double pi = acos(-1.0);const double eps = 1e-6;const int maxn = 1e6 + 5;const int maxm = 1e6 + 5;const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};inline int scan(int &a) { return scanf("%d", &a); }inline int scan(int &a, int &b) { return scanf("%d%d", &a, &b); }inline int scan(int &a, int &b, int &c) { return scanf("%d%d%d", &a, &b, &c); }inline int scan(ll &a) { return scanf("%I64d", &a); }inline int scan(ll &a, ll &b) { return scanf("%I64d%I64d", &a, &b); }inline int scan(ll &a, ll &b, ll &c) { return scanf("%I64d%I64d%I64d", &a, &b, &c); }inline int scan(double &a) { return scanf("%lf", &a); }inline int scan(double &a, double &b) { return scanf("%lf%lf", &a, &b); }inline int scan(double &a, double &b, double &c) { return scanf("%lf%lf%lf", &a, &b, &c); }inline int scan(char &a) { return scanf("%c", &a); }inline int scan(char *a) { return scanf("%s", a); }template<class T> inline void mem(T &A, int x) { memset(A, x, sizeof(A)); }template<class T0, class T1> inline void mem(T0 &A0, T1 &A1, int x) { mem(A0, x), mem(A1, x); }template<class T0, class T1, class T2> inline void mem(T0 &A0, T1 &A1, T2 &A2, int x) { mem(A0, x), mem(A1, x), mem(A2, x); }template<class T0, class T1, class T2, class T3> inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, int x) { mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x); }template<class T0, class T1, class T2, class T3, class T4> inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, int x) { mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x); }template<class T0, class T1, class T2, class T3, class T4, class T5> inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, int x) { mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x), mem(A5, x); }template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6, int x) { mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x), mem(A5, x), mem(A6, x); }template<class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }template<class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }template<class T> inline T min(T a, T b, T c, T d) { return min(min(a, b), min(c, d)); }template<class T> inline T max(T a, T b, T c, T d) { return max(max(a, b), max(c, d)); }template<class T> inline T min(T a, T b, T c, T d, T e) { return min(min(min(a,b),min(c,d)),e); }template<class T> inline T max(T a, T b, T c, T d, T e) { return max(max(max(a,b),max(c,d)),e); }template<class T> inline void dbg(T a[], int n) { rep(i, 0, n) cout << a[i] << (i == n - 1 ? "\n" : " ");}template<class T> inline void dbg(T a) { cout << a << " "; }template<class T> inline void dbg(T a[][maxn], int n, int m) { rep(i, 0, n) rep(j, 0, m) cout << a[i][j] << (j == m - 1 ? "\n" : " "); }