codeforces 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 = , f = ; char ch = getchar();

    while(ch < '' || ch > '') {

        if(ch == '-')f = -;

        ch = getchar();

    }

    while(ch >= '' && ch <= '') {

        x = x *  + ch - '';

        ch = getchar();

    }

    return x * f;

}

const double eps = 1e-;

const int mod = 1e9 + ;

const int maxn = 2e5 + ;

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 = ; i < n; i++) {

        scanf("%d%d", &l[i], &r[i]);

    }

    scanf("%d", &k);

    for(int i = ; i < n; i++) {

        if(l[i] <= k && k <= r[i]) {

            cout << n - i << endl;

            return ;

        }

    }

    return ;

}

codeforces 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 = , f = ; char ch = getchar();

    while(ch < '' || ch > '') {

        if(ch == '-')f = -;

        ch = getchar();

    }

    while(ch >= '' && ch <= '') {

        x = x *  + ch - '';

        ch = getchar();

    }

    return x * f;

}

const double eps = 1e-;

const int mod = 1e9 + ;

const int maxn = 2e5 + ;

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 - , n - k) +  * n << endl;

    return ;

}

codeforces 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 = , f = ; char ch = getchar();

    while(ch < '' || ch > '') {

        if(ch == '-')f = -;

        ch = getchar();

    }

    while(ch >= '' && ch <= '') {

        x = x *  + ch - '';

        ch = getchar();

    }

    return x * f;

}

const double eps = 1e-;

const int mod = 1e9 + ;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const LL INFLL = 0x3f3f3f3f3f3f3f3f;

int a[][];

int b[][];

vector<int> vec1[], vec2[];

int main() {

#ifndef ONLINE_JUDGE

    FIN

#endif

    int n, m;

    scanf("%d%d", &n, &m);

    for(int i = ; i < n; i++) {

        for(int j = ; j < m; j++) {

            scanf("%d", &a[i][j]);

        }

    }

    for(int i = ; i < n; i++) {

        for(int j = ; j < m; j++) {

            scanf("%d", &b[i][j]);

        }

    }

    for(int i = ; i < n; i++) {

        for(int j = ; j < m; j++) {

            vec1[i + j].push_back(a[i][j]);

            vec2[i + j].push_back(b[i][j]);

        }

    }

    for(int i = ; i < n + m - ; i++) {

        sort(vec1[i].begin(), vec1[i].end());

        sort(vec2[i].begin(), vec2[i].end());

        if(vec1[i] != vec2[i]) {

            cout << "NO" << endl;

            return ;

        }

    }

    cout << "YES" << endl;

    return ;

}

codeforces 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 = , f = ; char ch = getchar();

    while(ch < '' || ch > '') {

        if(ch == '-')f = -;

        ch = getchar();

    }

    while(ch >= '' && ch <= '') {

        x = x *  + ch - '';

        ch = getchar();

    }

    return x * f;

}

const double eps = 1e-;

const int mod = 1e9 + ;

const int maxn = 2e6 + ;

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 = ; i <= n; i++) {

        scanf("%d", &a[i]);

    }

    vec.push_back(a[n]);

    for(int i = , u, v; i < m; i++) {

        scanf("%d%d", &u, &v);

        s.insert(make_pair(u, v));

    }

    int ans = ;

    for(int i = n - ; i >= ; i--) {

        int flag = ;

        for(int j = ; j < vec.size(); j++) {

            if(s.count(make_pair(a[i], vec[j])) == ) {

                flag = ;

                break;

            }

        }

        if(flag) {

            ans++;

        } else {

            vec.push_back(a[i]);

        }

    }

    cout << ans << endl;

    return ;

}

codeforces 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 = , f = ; char ch = getchar();

    while(ch < '' || ch > '') {

        if(ch == '-')f = -;

        ch = getchar();

    }

    while(ch >= '' && ch <= '') {

        x = x *  + ch - '';

        ch = getchar();

    }

    return x * f;

}

const int maxn =  2e5 + ;

LL sum[maxn << ];

LL lazy[maxn << ];

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 << ] + sum[rt <<  | ];

}

void push_down(int l, int r, int rt) {

    if(lazy[rt] == NUL) return;

    int mid = (l + r) >> ;

    lazy[rt << ] = lazy[rt];

    lazy[rt <<  | ] = lazy[rt];

    sum[rt << ] = (LL)(mid - l + ) * lazy[rt] + c[mid + ] - c[l];

    sum[rt <<  | ] = (LL)(r - mid) * lazy[rt] + c[r + ] - c[mid + ];

    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) >> ;

    build(l, mid, rt << );

    build(mid + , r, rt <<  | );

    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 + ) * val + c[r + ] - c[l];

        lazy[rt] = val;

        return;

    }

    int mid = (l + r) >> ;

    push_down(l, r, rt);

    if(L <= mid) update(L, R, val, l, mid, rt << );

    if(R > mid) update(L, R, val, mid + , r, rt <<  | );

    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) >> ;

    push_down(l, r, rt);

    LL ans = ;

    if(L <= mid)  ans += query(L, R, l, mid, rt << );

    if(R > mid) ans += query(L, R, mid + , r, rt <<  | );

    return ans;

}

int main() {

#ifndef ONLINE_JUDGE

    FIN

#endif

    int n;

    scanf("%d", &n);

    for(int i = ; i < n; i++) {

        scanf("%lld", &a[i]);

    }

    for(int i = ; i < n; i++) {

        scanf("%lld", &k[i]);

        k[i] = k[i] + k[i - ];

    }

    for(int i = ; i <= n; i++) {

        c[i] = c[i - ] + k[i - ];

    }

    build(, n - , );

    int q;

    cin >> q;

    while(q--) {

        char op[];

        int l, r, pos;

        LL val;

        cin >> op;

        if(op[] == '+') {

            cin >> pos >> val;

            pos--;

            int l = pos;

            int r = n - ;

            LL tmp = query(pos, pos, , n - , );

            while(l < r) {

                int mid = (l + r + ) / ;

                if(tmp + val + k[mid] - k[pos] > query(mid, mid, , n - , )) {

                    l = mid;

                } else {

                    r = mid - ;

                }

            }

            // debug3(pos, r, tmp + val - k[pos]);

            update(pos, r, tmp + val - k[pos], , n - , );

        } else {

            cin >> l >> r;

            cout << query(l - , r - , , n - , ) << endl;;

            // printf("%lld\n", query(l - 1, r - 1, 0, n - 1, 1));

        }

    }

}

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