Codeforces Round #704 (Div. 2) A-E题解

A Three swimmers
题意三个人每人游一个来回时间分别是a、b、c，那么在 a、b、c的倍数时间点上三个人均会在左边的点，题目问你p时刻来还要等多久最快遇到三个人 1e18 除法判断是否整除相减即可

#include<cstdio>
#include<vector>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
const int MAX_N = 100005;
priority_queue<pair<int,int> > q;
int ans[MAX_N],arr[MAX_N];
int num = 0;
int main()
{int t;scanf("%d",&t);while(t--){num = 0;int n;scanf("%d",&n);for(int i = 1;i<=n;++i){scanf("%d",&arr[i]);q.push(make_pair(arr[i],i));}int maxx = n;while(!q.empty()){pair<int,int> top = q.top();q.pop();if(top.second>maxx) continue;int xb = top.second;while(xb<=maxx){ans[++num] = arr[xb];xb++;}maxx = top.second-1;}for(int i = 1;i<=n;++i){i==n?printf("%d\n",ans[i]):printf("%d ",ans[i]);}}return 0;
}

BCard Deck
按照卡牌顺序你每次选k张，然后将k张从最下面反过来放到新桌子上获得收益，收益因为是n的次方，我们发现每个卡牌是小于等于n，所以你把大的放在最下面获得收益最高，如果每个卡牌不是小于等于n 需要dp计算

#include<cstdio>
#include<vector>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
const int MAX_N = 100005;
priority_queue<pair<int,int> > q;
int ans[MAX_N],arr[MAX_N];
int num = 0;
int main()
{int t;scanf("%d",&t);while(t--){num = 0;int n;scanf("%d",&n);for(int i = 1;i<=n;++i){scanf("%d",&arr[i]);q.push(make_pair(arr[i],i));}int maxx = n;while(!q.empty()){pair<int,int> top = q.top();q.pop();if(top.second>maxx) continue;int xb = top.second;while(xb<=maxx){ans[++num] = arr[xb];xb++;}maxx = top.second-1;}for(int i = 1;i<=n;++i){i==n?printf("%d\n",ans[i]):printf("%d ",ans[i]);}}return 0;
}

C. Maximum width
定义完美字符串为T每个字符按顺序均属于S 两个字符间最长距离为所求值，一开始想到可能和长度有关压位dp，但是不太好定义于是定义出如下dp含义 dp[i][2] - dp[i][0] 代表T中第 i 个字符属于S最左能在什么位置当然要比dp[i-1][0] 大，那么dp[i][1] 代表T中第 i 个字符属于S最右能在什么位置当然要比 dp[i+1][1] 小枚举点求答案

#include<cstdio>
#include<vector>
#include<cstring>
#include<cmath>
#include<iostream>
#include<queue>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
const int MAX_N = 200005;
priority_queue<pair<int,int> > q;
long long dp[MAX_N][2];
char str[MAX_N],str_[MAX_N];
vector<int> vt[35];
int main()
{int n,m,ans = 0;scanf("%d%d",&n,&m);scanf("%s",str+1);for(int i = 1;i<=n;++i){vt[str[i]-'a'+1].push_back(i);}scanf("%s",str_+1);for(int i = 1;i<=m;++i){dp[i][0] = 0x3f3f3f3f;dp[i][1] = 0;}dp[0][0] = 0,dp[m+1][1] = 0x3f3f3f3f;for(int i = 1;i<=m;++i){int l = 0,r = vt[str_[i]-'a'+1].size()-1;while(l<=r){int mid = (l+r)/2;if(vt[str_[i]-'a'+1][mid] > dp[i-1][0]) r = mid - 1;else l = mid + 1;}dp[i][0] = vt[str_[i]-'a'+1][l];}for(int i = m;i>=1;--i){int l = 0,r = vt[str_[i]-'a'+1].size()-1;while(l<=r){int mid = (l+r)/2;if(vt[str_[i]-'a'+1][mid] < dp[i+1][1]) l = mid+1;else r = mid - 1;}dp[i][1] = vt[str_[i]-'a'+1][r];}dp[m+1][1] = 0;for(int i = 1;i<=m;++i){ans = max(1ll*ans,dp[i+1][1]-dp[i][0]);}printf("%d\n",ans);return 0;
}

D. Genius’s Gambit
当a=0或者b=1情况比较明显
手完出了 k<=a 与 k <=b-1的情况，写了个暴力归纳k<=a+b-2的情况 a+b-2可以将 k<=a 与 k<=b-1结合起来玩

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int MAX_N = 200025;
char str[MAX_N];
int main()
{int a,b,k;scanf("%d%d%d",&a,&b,&k);if(a==0){if(k==0){printf("Yes\n");for(int i = 1;i<=b;++i) printf("1");printf("\n");for(int i = 1;i<=b;++i) printf("1");printf("\n");}else{printf("No\n");}}else if(b==1){if(k==0){printf("Yes\n");for(int i = 1;i<=b;++i) printf("1");for(int i = 1;i<=a;++i) printf("0");printf("\n");for(int i = 1;i<=b;++i) printf("1");for(int i = 1;i<=a;++i) printf("0");printf("\n");}else{printf("No\n");}}else{if(k<=a){printf("Yes\n");for(int i = 1;i<=b-1;++i) printf("1");printf("1");for(int i = 1;i<=k;++i) printf("0");for(int i = 1;i<=a-k;++i) printf("0");printf("\n");for(int i = 1;i<=b-1;++i) printf("1");for(int i = 1;i<=k;++i) printf("0");printf("1");for(int i = 1;i<=a-k;++i) printf("0");printf("\n");}else if(k<=b-1){printf("Yes\n");printf("1");for(int i = 1;i<=a-1;++i) printf("0");for(int i = 1;i<=k;++i) printf("1");printf("0");for(int i = k+2;i<=b;++i) printf("1");printf("\n");printf("1");for(int i = 1;i<=a-1;++i) printf("0");printf("0");for(int i = 1;i<=k;++i) printf("1");for(int i = k+2;i<=b;++i) printf("1");printf("\n");}else if(k<=a+b-2){int len = 0;printf("Yes\n");for(int i = 1;i<=b;++i) printf("1");for(int i = 1;i<=a;++i) printf("0");printf("\n");for(int i = 1;i<=b-1;++i) str[++len] = '1';int xb = b;for(int i = 1;i<=a;++i) str[++len] = '0';str[++len] = '1';for(int i = 1;i<=k-a;++i) swap(str[xb],str[xb-1]),xb--;printf("%s\n",str+1);}else{printf("No\n");}}return 0;
}

E. Almost Fault-Tolerant Database
这题给你 n∗mn*mn∗m 个数，问你能否找到一行 mmm 个数，使得这 mmm 个数与每一行不同的数不超过 222 个
我们先考虑答案肯定和第一行不超过两个数（如果存在）那么我们去拿第一行假设作为答案与其余行求 maxdiffmaxdiffmaxdiff 如果 maxdiffmaxdiffmaxdiff >= 555 肯定无解，因为你只能换两个位置 5−2=3>=25-2=3 >= 25−2=3>=2 那么当 maxdiff<=2maxdiff <= 2maxdiff<=2 肯定输出第一行本身如果 maxdiff=4maxdiff = 4maxdiff=4 则交换其中两个元素， maxdiff=3maxdiff=3maxdiff=3 比较麻烦，先交换一个，然后看是否满足情况如果继续 maxdiff=3maxdiff = 3maxdiff=3 则做组合交换 3∗33*33∗3 情况，题解说 333 种情况可能还是有不同的地方另外代码也写的不够美观感觉定义的太随意了代码就越来越臃肿

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<stack>
#include<cstring>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
using namespace std;#define ll long longconst int MAX_N = 250025;
vector<int> vt[MAX_N];
int ans[MAX_N],cnt[MAX_N],maxx,xb,n,m,tmp[5],tmpLen,maxx_,xb_,tmp_[5],tmpLen_,maxx__,xb__;void findDiff()
{maxx = -1;for(int i = 1;i<=n;++i) cnt[i] = 0;for(int i = 1;i<=n;++i){for(int j = 1;j<=m;++j){if(vt[i][j]!=ans[j]) cnt[i]++;}if(cnt[i]>maxx){maxx = cnt[i];xb = i;}}
}void findDiffOne()
{maxx_ = -1;for(int i = 1;i<=n;++i) cnt[i] = 0;for(int i = 1;i<=n;++i){for(int j = 1;j<=m;++j){if(vt[i][j]!=ans[j]) cnt[i]++;}if(cnt[i]>maxx_){maxx_ = cnt[i];xb_ = i;}}
}void findDiffTwo()
{maxx__ = -1;for(int i = 1;i<=n;++i) cnt[i] = 0;for(int i = 1;i<=n;++i){for(int j = 1;j<=m;++j){if(vt[i][j]!=ans[j]) cnt[i]++;}if(cnt[i]>maxx__){maxx__ = cnt[i];xb__ = i;}}
}void GOOD()
{printf("Yes\n");for(int i = 1;i<=m;++i){i==m?printf("%d\n",ans[i]):printf("%d ",ans[i]);}
}bool gao(int x,int y,int initXb)
{for(int i = 1;i<=m;++i) ans[i] = vt[1][i];ans[tmp[x]] = vt[initXb][tmp[x]];ans[tmp[y]] = vt[initXb][tmp[y]];findDiff();if(maxx<=2){return true;}return false;
}bool gaoOne(int x)
{ans[tmp_[x]] = vt[xb_][tmp_[x]];findDiffTwo();if(maxx__<=2){return true;}return false;
}int main()
{int x;scanf("%d%d",&n,&m);for(int i = 1;i<=n;++i) vt[i].push_back(-1);for(int i = 1;i<=n;++i){for(int j = 1;j<=m;++j){scanf("%d",&x);vt[i].push_back(x);}}for(int i = 1;i<=m;++i) ans[i] = vt[1][i];findDiff();if(maxx<=2){GOOD();}else if(maxx>=5){printf("No\n");}else if(maxx==3){tmpLen = 0;for(int i = 1;i<=m;++i){if(ans[i]!=vt[xb][i]) tmp[++tmpLen] = i;}for(int i = 1;i<=tmpLen;++i){for(int j = 1;j<=m;++j) ans[j] = vt[1][j];ans[tmp[i]] = vt[xb][tmp[i]];findDiffOne();if(maxx_<=2){GOOD();return 0;}if(maxx_>3){continue;}tmpLen_ = 0;for(int j = 1;j<=m;++j) if(ans[j]!=vt[xb_][j]) tmp_[++tmpLen_] = j;for(int j = 1;j<=tmpLen_;++j){int lastOne = ans[tmp_[j]];if(gaoOne(j)){GOOD();return 0;}else{ans[tmp_[j]] = lastOne;}}}printf("No\n");}else{tmpLen = 0;for(int i = 1;i<=m;++i){if(ans[i]!=vt[xb][i]) tmp[++tmpLen] = i;}int initXb = xb;if(gao(1,2,initXb)||gao(1,3,initXb)||gao(1,4,initXb)||gao(2,3,initXb)||gao(2,4,initXb)||gao(3,4,initXb)){GOOD();}else{printf("No\n");}}return 0;
}

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