Codeforces Round #404 (Div. 2) 题解

题目链接：点击打开链接

这次比赛AC了4个水题， 然而我zz了Ｅ题写了个ｂｕｇ调了很久没时间写Ｄ啦。

A. Anton and Polyhedrons
水题， 加一加就行了。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;const int seed = 131;const ll INF64 = ll(1e18);const int maxn = 3000+10;int n;char s[100];int main() {    scanf("%d", &n);    int ans = 0;    for(int i = 1; i <= n; i++) {        scanf("%s", s);        if(s[0] == 'T') ans += 4;        else if(s[0] == 'C') ans += 6;        else if(s[0] == 'O') ans += 8;        else if(s[0] == 'D') ans += 12;        else ans += 20;    }    printf("%d\n", ans);    return 0;}

B. Anton and Classes
排序就行了， 我们肯定是在一个区间集合中找一个右端点最小的， 在另一个集合里找一个左端点最大的。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;const int seed = 131;const ll INF64 = ll(1e18);const int maxn = 200000+10;int n, m;struct node {    int l, r;    node(int l=0, int r=0):l(l), r(r) {}}a[maxn], b[maxn];bool cmp1(const node& a, const node& b) {    if(a.l != b.l) return a.l < b.l;    else return a.r < b.r;}bool cmp2(const node& a, const node& b) {    if(a.r != b.r) return a.r < b.r;    else return a.l < b.l;}int main() {    scanf("%d", &n);    for(int i = 1; i <= n; i++) {        scanf("%d%d", &a[i].l, &a[i].r);    }    scanf("%d", &m);    for(int i = 1; i <= m; i++) {        scanf("%d%d", &b[i].l, &b[i].r);    }    int ans = 0;    sort(a+1, a+1+n, cmp2);    sort(b+1, b+1+m, cmp1);    if(a[1].r <= b[m].l) {        ans = max(ans, b[m].l - a[1].r);    }    sort(a+1, a+1+n, cmp1);    sort(b+1, b+1+m, cmp2);    if(b[1].r <= a[n].l) {        ans = max(ans, a[n].l - b[1].r);    }    printf("%d\n", ans);    return 0;}

C. Anton and Fairy Tale
我们可以发现， 当ｎ＞ｍ时，　前ｍ天的每天早上一定会被加满谷仓，　在第ｍ＋１天开始，　下午的时候，　谷仓里谷子会少ｍ＋１个，ｍ＋２个．．．．，　所以我们只需要判断某一天ｃｎｔ，　当ｃｎｔ（ｃｎｔ＋１）／２　＋　ｍ　＞＝　ｎ，　我们可以发现，　这个是可以二分的，　二分他就行了。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;const int seed = 131;const ll INF64 = ll(1e18 + 10);const int maxn = 200000+10;ll n, m;bool ok(ll mid) {    ll cur = mid * (mid + 1) / 2;    return cur + m >= n;}int main() {    scanf("%I64d%I64d", &n, &m);    if(n <= m) {        printf("%I64d\n", n);        return 0;    }    ll l = 1, r = 2e9, ans = INF64;    while(l <= r) {        ll mid = (l + r) >> 1;        if(ok(mid)) {            r = mid - 1;            ans = min(ans, mid);        }        else l = mid + 1;    }    printf("%I64d\n", m + ans);    return 0;}

D. Anton and School - 2
Ｄ题其实很简单，　最后没时间写了，　我们如果从左向右枚举，　当枚举到一个（的时候，　我们假设一定选这个左括号，　那么我们要在之前的左括号里选０个，１个，２个．．．，　假设我们枚举到当前左括号，　左边有ｎ个，　右边有ｍ个右括号，　那么我们对答案贡献　ｓｕｍ（Ｃ（ｎ，０￣ｎ）×Ｃ（ｍ，１～ｎ＋１）＝　Ｃ（ｎ＋ｍ，　ｎ＋１）；

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int MOD = 1e9 + 7;const int maxn = 2e5 + 10;LL fac[maxn];void init(){    LL i;    fac[0]=1;    for (LL i = 1; i < maxn; i++)    fac[i] = fac[i - 1] * i % MOD;}LL exgcd(LL a, LL b, LL &x, LL &y) {    if (!b) {x = 1; y = 0; return a;}    LL d = exgcd(b, a % b, y, x);    y -= a / b * x;    return d;}LL inv(LL a, LL n) {    LL x, y;    exgcd(a, n, x, y);    return (x + n) % n;}LL C(LL n, LL m) {    return fac[n] * inv(fac[m] * fac[n - m] % MOD, MOD) % MOD;}char s[maxn];int main() {    init();    scanf("%s", s + 1);    int len = strlen(s+1);    int cnt1 = 0, cnt2 = 0;    LL ans = 0;    for(int i = 1; i <= len; i++) {        if(s[i] == ')') ++cnt2;    }    for(int i = 1; i <= len; i++) {//        printf("case :%d\n", i);        if(cnt2 == 0) break;        if(s[i] == '(') {//            printf("%d %d=", cnt1, cnt2);        ans = (ans + C(cnt1+cnt2, cnt1+1)) % MOD;           // ans = (ans + cnt2) % MOD;//            printf("%d\n", C(cnt1+cnt2, min(cnt1, cnt2)));            ++cnt1;        }        else --cnt2;//        printf("%d\n", ans);    }    printf("%I64d\n", ans);    return 0;}

E. Anton and Permutation
这种xjb更新，xjb询问的，我们就分块+二分暴力去搞就行啦。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <string>#include <vector>#include <stack>#include <bitset>#include <cstdlib>#include <cmath>#include <set>#include <list>#include <deque>#include <map>#include <queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;typedef long double ld;const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 0x3f3f3f3f;const int seed = 131;const ll INF64 = ll(1e18 + 10);const int maxn = 200000+10;int n, q;int belong[maxn];       /// ÿ���������ĸ���int block;              /// ��Ĵ�Сint num;                /// ��ĸ���int r[maxn], l[maxn];   /// ÿ������ұ߽�ll sum[maxn];           /// ����ά������Ϣll cnt[maxn];int a[maxn];vector<int> res[maxn/10];void build() {    block = sqrt(n);    num = n / block;    if(n % block) num++;    for(int i = 1; i <= num; i++) {        l[i] = (i-1)*block + 1;        r[i] = i * block;        sum[i] = 0;        cnt[i] = 0;    }    r[num] = n;    for(int i = 1; i <= n; i++) belong[i] = (i-1)/block+1;    int ccc = 1;    for(int i = 1; i <= num; i++) {        for(int j = l[i]; j <= r[i]; j++) {            res[i].push_back(ccc);            a[ccc] = ccc;            ++ccc;        }    }}int query(int x, int y, int cal) {    int ans = 0;    if(belong[x] == belong[y]) {        for(int i = x; i <= y; i++) {            if(a[i] < cal) ++ans;        }        return ans;    }    else {        for(int i = x; i <= r[belong[x]]; i++) {            if(a[i] < cal) ++ans;        }        for(int i = belong[x]+1; i < belong[y]; i++) {            int pos = lower_bound(res[i].begin(), res[i].end(), cal) - res[i].begin();            if(pos == (int)res[i].size()) ans += pos;            else if(res[i][pos] != cal) ans += pos;        }        for(int i = l[belong[y]]; i <= y; i++) {            if(a[i] < cal) ++ans;        }        return ans;    }}void update(int l, int r) {    int be = belong[l];    int bee = belong[r];    int pos = lower_bound(res[be].begin(), res[be].end(), a[l]) - res[be].begin();    res[be].erase(res[be].begin() + pos);    res[be].push_back(a[r]);    sort(res[be].begin(), res[be].end());    pos = lower_bound(res[bee].begin(), res[bee].end(), a[r]) - res[bee].begin();    res[bee].erase(res[bee].begin() + pos);    res[bee].push_back(a[l]);    sort(res[bee].begin(), res[bee].end());}void print() {    for(int i = 1; i <= num; i++) {        for(int j = l[i]; j <= r[i]; j++) {            printf("%d ", res[i][j-l[i]]);        }        cout<<endl;    }}int main() {    scanf("%d%d", &n, &q);    build();    ll ans = 0;    while(q--) {        int l, r;        scanf("%d%d", &l, &r);        if(l > r) swap(l, r);        if(l == r) {            printf("%I64d\n", ans);            continue;        }        if(l+1 == r) {            if(a[l] < a[r]) ++ans;            else --ans;        }        else {            int cur = query(l+1, r-1, a[l]);            int maxv = (r - l - 1) - cur;            ans -= cur;            ans += maxv;            cur = query(l+1, r-1, a[r]);            maxv = (r - l - 1) - cur;            ans += cur;            ans -= maxv;            if(a[l] < a[r]) ++ans;            else --ans;        }        update(l, r);        swap(a[l], a[r]);      //  print();        printf("%I64d\n", ans);    }    return 0;}