【AtCoder】AGC015

時間  2019-12-06
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AGC015

A - A+...+B Problem

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 300005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int64 N,A,B;
void Solve() {
    read(N);read(A);read(B);
    if(A > B) {puts("0");return;}
    if(A == B) {puts("1");return;}
    if(N == 1) {puts("0");return;}
    out(A + B * (N - 1) - B - A * (N - 1) + 1);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

B - Evilator

若是是U的話下面的層須要坐兩次，D的話上面的層須要坐兩次c++

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
char s[MAXN];
int64 ans;
int N;
void Solve() {
    scanf("%s",s + 1);
    N = strlen(s + 1);
    for(int i = 1 ; i <= N ; ++i) {
    if(s[i] == 'U') ans += N - i + (i - 1) * 2;
    else ans += i - 1 + (N - i) * 2;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

C - Nuske vs Phantom Thnook

由於造成一個森林，咱們只須要查詢這個邊框裏面有多少個樹根以及斷開了多少父親與兒子之間的邊spa

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M,Q;
char s[2005][2005];
int ch[4][2005][2005],sum[2005][2005],fa[2005 * 2005];
int dx[4] = {0,1,0,-1},dy[4] = {1,0,-1,0};
int id(int x,int y) {
    return (x - 1) * M + y;
}
int getfa(int u) {
    return fa[u] == u ? u : fa[u] = getfa(fa[u]);
}
void dfs(int x,int y) {
    for(int k = 0 ; k < 4 ; ++k) {
    int tx = x + dx[k],ty = y + dy[k];
    if(s[tx][ty] == '1') {
        if(fa[id(x,y)] != id(tx,ty)) {
        ch[k][tx][ty] = 1;
        fa[id(tx,ty)] = id(x,y);
        dfs(tx,ty);
        }
    }
    }
}
void Solve() {
    read(N);read(M);read(Q);
    for(int i = 1 ; i <= N ; ++i) {
    scanf("%s",s[i] + 1);
    }
    for(int i = 1 ; i <= N ; ++i) {
    for(int j = 1 ; j <= M ; ++j) {
        if(s[i][j] == '1' && !fa[id(i,j)]) {
        sum[i][j] = 1;
        dfs(i,j);
        }
    }
    }
    for(int i = 1 ; i <= N ; ++i) {
    for(int j = 1 ; j <= M ; ++j) {
        sum[i][j] += sum[i][j - 1];
    }
    }
    for(int i = 1 ; i <= N ; ++i) {
    for(int j = 1 ; j <= M ; ++j) {
        sum[i][j] += sum[i - 1][j];
    }
    }
    for(int j = 1 ; j <= M ; ++j) {
    for(int i = 1 ; i <= N ; ++i) {
        ch[0][i][j] += ch[0][i - 1][j];
        ch[2][i][j] += ch[2][i - 1][j];
    }
    }
    for(int i = 1 ; i <= N ; ++i) {
    for(int j = 1 ; j <= M ; ++j) {
        ch[1][i][j] += ch[1][i][j - 1];
        ch[3][i][j] += ch[3][i][j - 1];
    }
    }
    int x[2],y[2];
    for(int i = 1 ; i <= Q ; ++i) {
    read(x[0]);read(y[0]);read(x[1]);read(y[1]);
    int ans = 0;
    ans += ch[0][x[1]][y[0]] - ch[0][x[0] - 1][y[0]];
    ans += ch[1][x[0]][y[1]] - ch[1][x[0]][y[0] - 1];
    ans += ch[2][x[1]][y[1]] - ch[2][x[0] - 1][y[1]];
    ans += ch[3][x[1]][y[1]] - ch[3][x[1]][y[0] - 1];
    ans += sum[x[1]][y[1]] - sum[x[0] - 1][y[1]] - sum[x[1]][y[0] - 1] + sum[x[0] - 1][y[0] - 1]; 
    out(ans);enter;
    }
    
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - A or...or B Problem

找到\(A\)和\(B\)第一個不一樣的最高位是\(T = 2^{r}\)code

而後只保留\(A\)和\(B\)的後\(r\)位排序

分紅兩個集合get

\(\{A,A + 1....T - 1\}\)和\(\{T,T + 1...B\}\)it

第一個集合能獲得的數的區間是\([A,T - 1]\)class

第二個集合除了第\(r\)位能獲得的最高位是\(k\),能獲得的區間爲\([T,2^{r} + 2^{k + 1} - 1]\)date

兩個集合加一塊兒能獲得的區間是\([T + A,2T - 1]\)gc

區間求並便可統計

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
pair<int64,int64> p[5];
void Solve() {
    int64 A,B;
    read(A);read(B);
    if(A == B) {puts("1");return;}
    int r = -1;
    for(int i = 59 ; i >= 0 ; --i) {
    if((A >> i & 1) != (B >> i & 1)) {r = i;break;}
    }
    int64 ans = 0;
    A &= (1LL << r + 1) - 1;
    B &= (1LL << r + 1) - 1;
    p[1] = mp((1LL << r) + A,(1LL << r + 1) - 1);
    int t = -1;
    for(int i = r - 1 ; i >= 0 ; --i) {
    if(B >> i & 1) {t = i;break;}
    }
    p[0] = mp(A,(1LL << r) + (1LL << t + 1) - 1);
    sort(p,p + 2);
    int64 rm = -1;
    for(int i = 0 ; i < 2 ; ++i) {
    if(p[i].se < rm) continue;
    else {
        ans += p[i].se - max(rm,p[i].fi - 1);
        rm = p[i].se;
    }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Mr.Aoki Incubator

按照速度排序的話，若是選擇一個點，能影響到的點是座標值大於它且速度最小的到座標值小於它且速度最大的點這個區間

這些區間排序後左右端點不降

而後能夠用一個dp+前綴和統計一下覆蓋全局的方案數

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N;
pii p[MAXN],rg[MAXN];
int L[MAXN],R[MAXN],st[MAXN],top;
int dp[MAXN],s[MAXN];
vector<int> v[MAXN];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    read(N);
    int x,v;
    for(int i = 1 ; i <= N ; ++i) {
    read(x);read(v);p[i] = mp(v,x);
    }
    sort(p + 1,p + N + 1);
    
    for(int i = 1 ; i <= N ; ++i) {
    rg[i].fi = i;
    if(!top || p[st[top]].se < p[i].se) st[++top] = i;
    if(top) {
        int l = 1,r = top;
        while(l < r) {
        int mid = (l + r) >> 1;
        if(p[st[mid]].se >= p[i].se) r = mid;
        else l = mid + 1;
        }
        rg[i].fi = st[l];
    }
    
    }
    top = 0;
    for(int i = N ; i >= 1 ; --i) {
    rg[i].se = i;
    if(!top || p[st[top]].se > p[i].se) st[++top] = i;
    if(top) {
        int l = 1,r = top;
        while(l < r) {
        int mid = (l + r) >> 1;
        if(p[st[mid]].se <= p[i].se) r = mid;
        else l = mid + 1;
        }
        rg[i].se = st[l];
    }
    
    }
    sort(rg + 1,rg + N + 1);
    s[0] = 1;dp[0] = 1;
    int k = 1;
    for(int i = 1 ; i <= N ; ++i) {
    while(k <= rg[i].se) {
        s[k] = inc(s[k - 1],dp[k]);
        ++k;
    }
    int t = s[rg[i].se];
    if(rg[i].fi != 1) update(t,MOD - s[rg[i].fi - 2]);
    update(dp[rg[i].se],t);
    
    s[rg[i].se] = inc(s[rg[i].se - 1],dp[rg[i].se]);
    }
    out(dp[N]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F結論題，看不懂，咕了

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