Codeforces Round #546 (Div. 2) ABCDE 题解

其他 2019-03-13 13:34:41 阅读次数: 0

1136A：

题意：一本书有n个章节，每个章节的分别在li到ri页，小明读完书后将书折在第k页，问还有多少章节没有读

题解：控制k在li~ri的范围内后输出n-i即可

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
int l[maxn];
int r[maxn];
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n, k;
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%d%d", &l[i], &r[i]);
    }
    scanf("%d", &k);
    for(int i = 0; i < n; i++) {
        if(l[i] <= k && k <= r[i]) {
            cout << n - i << endl;
            return 0;
        }
    }
    return 0;
}
View Code

1136B:
题意：有n个井盖，井盖上有一颗石头，你每次可以做以下三种操作中的一种

1.将石头丢在隔壁的井盖上

2.走向相邻的井盖

3.如果井盖上没有石头，将井盖打开

你初始时在第k个井盖上，问你最少需要多少次操作可以打开全部的井盖

题解：每个井盖都需要进行3次操作才能到这个井盖上，首先我们需要到这个井盖上，然后我们需要搬石头，再然后我们要打开井盖，但是我们会走重复的路径，所以要想操作数最少，我们应该走重复的路最少，即min（n-k，k-1）

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;

int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n,k;
    scanf("%d%d", &n, &k);
    cout << min(k - 1, n - k) + 3 * n << endl;
    return 0;
}
View Code

1136C：

题意：给你一个矩阵A和一个矩阵B，问你矩阵A和矩阵B是否是同类矩阵(转置矩阵也是同类的

题解：记录一下两个矩阵对角线上的元素是否相等即可

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
int a[505][505];
int b[505][505];
vector<int> vec1[1005], vec2[1005];
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            scanf("%d", &a[i][j]);
        }
    }
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            scanf("%d", &b[i][j]);
        }
    }
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            vec1[i + j].push_back(a[i][j]);
            vec2[i + j].push_back(b[i][j]);
        }
    }
    for(int i = 0; i < n + m - 1; i++) {
        sort(vec1[i].begin(), vec1[i].end());
        sort(vec2[i].begin(), vec2[i].end());
        if(vec1[i] != vec2[i]) {
            cout << "NO" << endl;
            return 0;
        }
    }
    cout << "YES" << endl;
    return 0;
}
View Code

1136D：

题意：有一队人，其中有n对人可以两两交换，你在队伍的尾部，问你你最多可以向前移动多少步

题解：用一个vector记录你可以交换的人，然后判断你是否能与vector里面的人进行交换，如果你可以与vector里面所有的人交换的话，你就可以移动一步

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 2e6 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
set<pair<int, int> > s;
int a[maxn];
vector<int> vec;
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    vec.push_back(a[n]);
    for(int i = 0, u, v; i < m; i++) {
        scanf("%d%d", &u, &v);
        s.insert(make_pair(u, v));
    }
    int ans = 0;
    for(int i = n - 1; i >= 1; i--) {
        int flag = 1;
        for(int j = 0; j < vec.size(); j++) {
            if(s.count(make_pair(a[i], vec[j])) == 0) {
                flag = 0;
                break;
            }
        }
        if(flag) {
            ans++;
        } else {
            vec.push_back(a[i]);
        }
    }

    cout << ans << endl;
    return 0;
}
View Code

1136E：

题意：给你一个长度为n的序列a和长度为n-1的序列k，序列a在任何时候都满足如下性质，a[i+1]>=ai+ki，如果更新后a[i+1]<ai+ki了，那么a[i+1]=ai+ki

现在给你q次操作

操作1：将位置为pos的元素+x

操作2：询问区间l，r的区间和

题解：非常明显的线段树题，我们不好维护的是，如果更新后，当前数字变大到不满足限制条件时，我后面的元素也要做出相应的更新

那么我们就将a序列先减去k序列，这样的a序列也是满足限制条件了，然后我们记录下k的前缀和的前缀和，避免询问时缺少k的贡献，

数学推导如下

序列a满足单调不减性，

则∑ai 同样满足单调不减性，

当我们对位置为pos的元素进行更新时，

如果后面的元素 a[pos+R]<a[pos]，则该元素要被覆盖，

所以我们二分右端点，将区间【pos，R】覆盖为a[pos]+x即可

为了避免重复计算ki对a的贡献所以我们覆盖区间时可以用如下技巧

用c来记录k的前缀和的前缀和

每次覆盖时，我们将区间【l，r】覆盖为(a[pos]+x-k[pos])*（r-l+1）+c[r+1]-c[l]

这样就不会使得k的贡献计算错了

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
LL read() {
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
const int maxn =  2e5 + 5;
LL sum[maxn << 2];
LL lazy[maxn << 2];
LL a[maxn];
LL k[maxn];
LL sub[maxn];
LL c[maxn];
LL NUL = -1e18;
void push_up(int rt) {
    sum[rt] = (LL)sum[rt << 1] + sum[rt << 1 | 1];
}
void push_down(int l, int r, int rt) {
    if(lazy[rt] == NUL) return;
    int mid = (l + r) >> 1;
    lazy[rt << 1] = lazy[rt];
    lazy[rt << 1 | 1] = lazy[rt];
    sum[rt << 1] = (LL)(mid - l + 1) * lazy[rt] + c[mid + 1] - c[l];
    sum[rt << 1 | 1] = (LL)(r - mid) * lazy[rt] + c[r + 1] - c[mid + 1];
    lazy[rt] = NUL;
}
void build(int l, int r, int rt) {
    lazy[rt] = NUL;
    if(l == r) {
        sum[rt] = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    push_up(rt);
}
void update(int L, int R, LL val, int l, int r, int rt) {
    if(L <= l && r <= R) {
        sum[rt] = (LL)(r - l + 1) * val + c[r + 1] - c[l];
        lazy[rt] = val;
        return;
    }
    int mid = (l + r) >> 1;
    push_down(l, r, rt);
    if(L <= mid) update(L, R, val, l, mid, rt << 1);
    if(R > mid) update(L, R, val, mid + 1, r, rt << 1 | 1);
    push_up(rt);
}
LL query(int L, int R, int l, int r, int rt) {
    if(L <= l && r <= R) {
        //debug2(rt,sum[rt]);
        return sum[rt];
    }
    int mid = (l + r) >> 1;
    push_down(l, r, rt);
    LL ans = 0;
    if(L <= mid)  ans += query(L, R, l, mid, rt << 1);
    if(R > mid) ans += query(L, R, mid + 1, r, rt << 1 | 1);
    return ans;
}
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%lld", &a[i]);
    }
    for(int i = 1; i < n; i++) {
        scanf("%lld", &k[i]);
        k[i] = k[i] + k[i - 1];
    }
    for(int i = 1; i <= n; i++) {
        c[i] = c[i - 1] + k[i - 1];
    }
    build(0, n - 1, 1);
    int q;
    cin >> q;

    while(q--) {
        char op[2];
        int l, r, pos;
        LL val;
        cin >> op;

        if(op[0] == '+') {
            cin >> pos >> val;
            pos--;
            int l = pos;
            int r = n - 1;
            LL tmp = query(pos, pos, 0, n - 1, 1);
            while(l < r) {
                int mid = (l + r + 1) / 2;
                if(tmp + val + k[mid] - k[pos] > query(mid, mid, 0, n - 1, 1)) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            // debug3(pos, r, tmp + val - k[pos]);
            update(pos, r, tmp + val - k[pos], 0, n - 1, 1);
        } else {
            cin >> l >> r;
            cout << query(l - 1, r - 1, 0, n - 1, 1) << endl;;
            // printf("%lld\n", query(l - 1, r - 1, 0, n - 1, 1));
        }
    }
}
View Code

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