Codeforces Round #574 (Div. 2)题解
比赛链接
A题
题意
\(n\)个人每个人都有自己喜欢喝的\(vechorka\)口味，现在给你\(\lceil n/2\rceil\)箱\(vechorka\)，每箱有两瓶，问最多能有多少个人能拿到自己喜欢的口味。
思路
我们首先记录每个口味有多少个人喜欢，然后要想拿到自己喜欢的口味最大那么一定要优先考虑能凑偶数的，把偶数考虑完后剩余的口味一定都是\(1\)，就不管怎么分都只能满足一半的人。
代码实现如下
#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 <cassert>
#include <iostream>
#include <algorithm>
#include <unordered_map>
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 rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
int n, k;
int a[maxn], num[maxn];
int main() {
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
num[a[i]]++;
}
int ans = 0;
int tot = (n + 1) / 2;
for(int i = 1; i <= k; ++i) {
ans += num[i] / 2 * 2;
tot -= num[i] / 2;
num[i] %= 2;
}
ans += tot;
printf("%d\n", ans);
return 0;
}
B题
题意
要你使用恰好\(n\)次操作使得总糖果数为\(k\)，操作分为两种：
- 增加上一次增加的数量\(+1\)个糖果；
- 减少\(1\)个糖果。
思路
二分\(check\)。
代码实现如下
#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 <cassert>
#include <iostream>
#include <algorithm>
#include <unordered_map>
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 rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
int n, k;
bool check(int x) {
return (1LL * x * x + 3LL * x) / 2 - n >= k;
}
int main() {
scanf("%d%d", &n, &k);
int ub = n, lb = 1, mid, ans = 0;
while(ub >= lb) {
mid = (ub + lb) >> 1;
if(check(mid)) {
ans = mid;
ub = mid - 1;
} else {
lb = mid + 1;
}
}
printf("%d\n", n - ans);
return 0;
}
C题
题意
总共有\(2n\)个人，第一排的编号从\(1\)到\(n\)，第二排也是，现在要你选择任意多个人使得总身高最大，但是注意同一个编号只能有一个人，编号相邻的话不能是同一排的。
思路
\(dp[i][j]\)表示编号为\(i\)的人选择状态为\(j\)时的最大身高，\(j=0\)表示从第一排选，\(j=1\)从第二排，\(j=2\)为不选，则\(dp[i][0]=max(dp[i-1][1],dp[i-1][2])+a[i],dp[i][1]=max(dp[i-1][0],dp[i-1][2])+b[i],dp[i][2]=max(dp[i-1][0],dp[i-1][1],dp[i-1][2])\)。
代码实现如下
#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 <cassert>
#include <iostream>
#include <algorithm>
#include <unordered_map>
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 rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
int n;
int a[maxn], b[maxn];
LL dp[maxn][3];
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
for(int i = 1; i <= n; ++i) {
scanf("%d", &b[i]);
}
for(int i = 1; i <= n; ++i) {
dp[i][0] = max(dp[i-1][1], dp[i-1][2]) + a[i];
dp[i][1] = max(dp[i-1][0], dp[i-1][2]) + b[i];
dp[i][2] = max(dp[i-1][0], max(dp[i-1][1], dp[i-1][2]));
}
printf("%lld\n", max(dp[n][0], max(dp[n][1], dp[n][2])));
return 0;
}
D题
题意
定义\(f\)函数为
如果\(p\geq q\)：\(f(a1…ap,b1…bq)=a_1a_2\dots a_{p−q+1}b_1a{p−q+2}b_2\dots a_{p−1}b_{q−1}a_pb_q\);
如果\(p<q\)：\(f(a_1\dots a_p,b_1\dots b_q)=b_1b_2…b_{q−p}a_1b_{q−p+1}a_2\dots a_{p−1}b{q−1}a_pb_q\).
思路
按位算贡献，先预处理出\(a_i\)与长度\(len\)的数进行\(f\)函数的贡献然后乘以长度为\(len\)的数的个数。
代码实现如下
#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 <cassert>
#include <iostream>
#include <algorithm>
#include <unordered_map>
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 rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
const double eps = 1e-8;
const int mod = 998244353;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
int n;
int a[maxn], pw[105], cnt[30];
LL dp[maxn][30];
int main() {
scanf("%d", &n);
pw[0] = 1;
for(int i = 1; i < 30; ++i) {
pw[i] = 1LL * pw[i-1] * 10 % mod;
}
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = 1; i <= n; ++i) {
int x = a[i], len = 0;
while(x) {
++len;
x /= 10;
}
cnt[len]++;
for(int j = 1; j <= 10; ++j) {
if(len >= j) {
int num = a[i];
for(int k = 0; k < j; ++k) {
int x = num % 10;
num /= 10;
dp[i][j] = (((dp[i][j] + 1LL * x * pw[k*2] % mod) % mod) + 1LL * x * pw[k*2+1] % mod) % mod;
}
int pp = 2 * j;;
for(int k = j; k < len; ++k) {
int x = num % 10;
num /= 10;
dp[i][j] = (dp[i][j] + 2LL * x * pw[pp] % mod) % mod;
++pp;
}
} else {
int num = a[i];
for(int k = 0; k < len; ++k) {
int x = num % 10;
num /= 10;
dp[i][j] = (((dp[i][j] + 1LL * x * pw[k*2] % mod) % mod) + 1LL * x * pw[k*2+1] % mod) % mod;
}
}
}
}
LL ans = 0;
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= 10; ++j) {
ans = (ans + 1LL * dp[i][j] * cnt[j] % mod) % mod;
}
}
printf("%lld\n", ans);
return 0;
}
E题
题意
给你构造\(n\times m\)的矩阵的公式然后要你求所有大小为\(a\times b\)的子矩阵内最小值的和。
思路
首先我们先通过暴力将矩阵构造出来，然后对每一行用单调队列求出最小值，然后再把这个值当成\(mp[i][j]\)再对每一列求一次然后累加即可。
代码实现如下
#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 <cassert>
#include <iostream>
#include <algorithm>
#include <unordered_map>
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 rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
const double eps = 1e-8;
const int mod = 998244353;
const int maxn = 1e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
int n, m, a, b;
LL g, x, y, z;
int mp[3002][3002], dp[3002][3002];
deque<pii> q;
int main() {
scanf("%d%d%d%d", &n, &m, &a, &b);;
scanf("%lld%lld%lld%lld", &g, &x, &y, &z);
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= m; ++j) {
mp[i][j] = g;
g = (1LL * g * x % z + y) % z;
}
}
for(int i = 1; i <= n; ++i) {
while(!q.empty()) q.pop_back();
for(int j = m; j >= 1; --j) {
while(!q.empty() && mp[i][j] < q.front().first) q.pop_front();
q.push_front({mp[i][j], j});
dp[i][j] = q.back().first;
while(!q.empty() && q.back().second >= j + b - 1) q.pop_back();
}
}
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= m; ++j) {
mp[i][j] = dp[i][j];
}
}
for(int j = 1; j <= m; ++j) {
while(!q.empty()) q.pop_back();
for(int i = n; i >= 1; --i) {
while(!q.empty() && mp[i][j] < q.front().first) q.pop_front();
q.push_front({mp[i][j], i});
dp[i][j] = q.back().first;
while(!q.empty() && q.back().second >= i + a - 1) q.pop_back();
}
}
LL ans = 0;
for(int i = 1; i <= n - a + 1; ++i) {
for(int j = 1; j <= m - b + 1; ++j) {
ans += dp[i][j];
}
}
printf("%lld\n", ans);
return 0;
}
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