A. Arithmetic Derivative

形如$p^p(p是質數)$的數的比值為$1$，用$k$個這種數相乘得到的數的比值為$k$，爆搜即可。

#include<cstdio> #include<algorithm> typedef unsigned long long ll; int K,cnt,all;ll n,i,q[100000],ans[1000000]; ll po(ll a,ll b){ll t=1;while(b--){if(t>n/a)return n+1;t*=a;}return t; } bool isprime(ll n){for(ll i=2;i<n;i++)if(n%i==0)return 0;return 1; } void dfs(int x,ll y,int k){if(y>n)return;//printf("%d %llu %d\n",x,y,k);if(k==K){ans[++all]=y;return;}if(x>cnt)return;dfs(x+1,y,k);if(y<=n/q[x])dfs(x,y*q[x],k+1); } int main(){scanf("%d%llu",&K,&n);for(i=2;;i++){if(po(i,i)>n)break;if(isprime(i))q[++cnt]=po(i,i);}dfs(1,1,0);printf("%d\n",all);std::sort(ans+1,ans+all+1);for(i=1;i<=all;i++)printf("%llu ",ans[i]); }

　　

B. White Triangle

留坑。

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C. New Street

用set維護相同連續段，每次新增貢獻時利用多項式求冪，刪除貢獻則采用多項式求逆。

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D. Clones and Treasures

貪心，每次不滿足條件了就把前面全部舍棄。

#include<stdio.h> #include<iostream> #include<string.h> #include<string> #include<ctype.h> #include<math.h> #include<set> #include<map> #include<vector> #include<queue> #include<bitset> #include<algorithm> #include<time.h> using namespace std; void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); } #define MS(x, y) memset(x, y, sizeof(x)) #define ls o<<1 #define rs o<<1|1 typedef long long LL; typedef unsigned long long UL; typedef unsigned int UI; template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; } template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; } const int N = 1e6 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f; template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; } int casenum, casei; char s[N]; int n; int main() {while(~scanf("%s", s + 1)){n = strlen(s + 1);int sum = 0, num = 0;int ans = 0;for(int i = 1; i <= n; ++i){int val;if(s[i] == 'H')val = 1;else val = (s[i] == 'M' ? 0 : -1);sum += val;num += (val == 0);if(sum < 0){sum = num = 0;}else{gmax(ans, num);}}printf("%d\n", ans);}return 0; } /* 【trick&&吐槽】【題意】【分析】【時間復雜度&&優化】*/

　　

E. Space Tourists

當$n>k$時答案為$k$。

否則最優策略是將$k$個數平均分成$n-1$組，每組$size(size-1)$個字符串，外加$k$個兩個字符相等的字符串。

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F. Matrix Game

最小表示法。

#include<stdio.h> #include<iostream> #include<string.h> #include<string> #include<ctype.h> #include<math.h> #include<set> #include<map> #include<vector> #include<queue> #include<bitset> #include<algorithm> #include<time.h> using namespace std; void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); } #define MS(x, y) memset(x, y, sizeof(x)) #define ls o<<1 #define rs o<<1|1 typedef long long LL; typedef unsigned long long UL; typedef unsigned int UI; template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; } template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; } const int N = 105, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f; template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; } int casenum, casei; int n, m; char s[N][N]; struct A {string s;bool operator < (const A & b)const{return s < b.s;} }line[N], list[N]; int main() {while(~scanf("%d%d", &n, &m)){for(int i = 1; i <= n; ++i){scanf("%s", s[i] + 1);line[i].s = s[i] + 1;}for(int i = 1; i <= m; ++i){string now = "";for(int j = 1; j <= n; ++j){now = now + s[j][i];}list[i].s = now;}sort(line + 1, line + n + 1);sort(list + 1, list + m + 1);string ans = "";for(int i = 1; i <= n; ++i){string body = "";for(int j = n; j >= 1; --j)if(j != i){body = body + line[j].s;}for(int j = 0; j < m; ++j){string tmp = line[i].s.substr(j)+body+line[i].s.substr(0, j);gmax(ans, tmp);}}for(int i = 1; i <= m; ++i){string body = "";for(int j = m; j >= 1; --j)if(j != i){body = body + list[j].s;}for(int j = 0; j < n; ++j){string tmp = list[i].s.substr(j) + body + list[i].s.substr(0, j);gmax(ans, tmp);}}int g = ans.size();int st = g - 1;for(int i = 0; i < g; ++i)if(ans[i] != '0'){st = i;break;}cout << ans.substr(st) << endl;}return 0; } /* 【trick&&吐槽】【題意】【分析】【時間復雜度&&優化】 2 3 000 001*/

　　

G. Milkland

留坑。

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H. Parallel Relay

假如知道了所有數，那么可以等效成只有最小值和最大值的情況。

對于每一段，將所有數翻折到左邊后排序，那么最小值和最大值分別是從左往右的第一個未翻折到右側和翻折到右側的，共$O(n)$種情況，枚舉即可。

時間復雜度$O(n\log n)$。

#include<cstdio> #include<algorithm> using namespace std; const int N=100010; int n,m,i,j,a[N],b[N],c[N],ans; inline int cal(int a,int b,int c,int d){int ret=~0U>>1;for(int i=0;i<2;i++){swap(c,d);ret=min(ret,abs(a-c)+abs(c-d)+abs(d-b));}return ret; } int solve(int L,int R){int i,j;int cnt=0;for(i=L;i<=R;i++){c[++cnt]=b[i];}for(i=2;i<cnt;i++)c[i]=min(c[i],m-c[i]+1);sort(c+2,c+cnt);if(cnt==2)return abs(c[1]-c[2]);if(cnt==3)return min(abs(c[1]-c[2])+abs(c[2]-c[3]),abs(c[1]-(m-c[2]+1))+abs(m-c[2]+1-c[3]));int ret=cal(c[1],c[cnt],c[2],c[cnt-1]);ret=min(ret,cal(c[1],c[cnt],m-c[2]+1,m-c[cnt-1]+1));for(i=3;i<cnt;i++){ret=min(ret,cal(c[1],c[cnt],c[2],m-c[i]+1));ret=min(ret,cal(c[1],c[cnt],m-c[2]+1,c[i]));}return ret; } int main(){scanf("%d%d",&n,&m);n++;scanf("%d%d",&b[1],&b[n]);a[1]=a[n]=1;for(i=2;i<n;i++)scanf("%d%d",&a[i],&b[i]);for(i=1,j=0;i<=n;i++)if(a[i]==1){if(j)ans+=solve(j,i);j=i;}printf("%d",ans); } /* 5 8 2 6 1 7 2 3 1 1 2 15 21 4 15 2 5 2 6 1 2 2 210 10 1 10 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 */

　　

I. Minimum Prefix

二分答案，然后Hopcroft求二分圖最大匹配檢驗。

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J. Terminal

$f[i][j]$表示考慮前$i$組，第一輛車上了$j$個人是否可行。

#include<stdio.h> #include<iostream> #include<string.h> #include<string> #include<ctype.h> #include<math.h> #include<set> #include<map> #include<vector> #include<queue> #include<bitset> #include<algorithm> #include<time.h> using namespace std; void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); } #define MS(x, y) memset(x, y, sizeof(x)) #define ls o<<1 #define rs o<<1|1 typedef long long LL; typedef unsigned long long UL; typedef unsigned int UI; template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; } template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; } const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f; template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; } int casenum, casei; int n, m, K; int a[N]; int cnt[N], tmp[N]; bitset<N>f[2020]; int main() {while(~scanf("%d%d%d", &n, &m, &K)){gmin(K, n);MS(cnt, 0);MS(tmp, 0);for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);++cnt[a[i]];}f[0][0] = 1;int g = 0;LL ans = 1e18;for(int i = 1; i <= n; ++i){++tmp[a[i]];if(tmp[a[i]] == cnt[a[i]]){++g;f[g] = f[g - 1] | (f[g - 1] << cnt[a[i]]);for(int j = 1; j <= K; ++j)if(f[g][j] && n - j <= K){gmin(ans, (LL)i * j + (LL)n * (n - j));}}}if(ans == 1e18)ans = -1;printf("%lld\n", ans);}return 0; } /* 【trick&&吐槽】【題意】【分析】【時間復雜度&&優化】*/

　　

K. New Tetris

用線段樹加速模擬即可。

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L. Canonical duel

每個炮塔可以將其同行同列的所有炮塔都激活，因此枚舉一個點，然后將上下左右兩個連通塊合并即可。

#include<cstdio> const int N=2010,M=N*N; int n,m,i,j,k,id[N][N],cnt,f[M],size[M]; int ans=-1,px,py; char a[N][N]; int left[N][N],right[N][N],up[N][N],down[N][N]; int F(int x){return f[x]==x?x:f[x]=F(f[x]);} inline void merge(int x,int y){x=F(x),y=F(y);if(x==y)return;f[x]=y; } inline void gao(int x,int y){int A=up[x][y];if(!A)A=down[x][y];int B=left[x][y];if(!B)B=right[x][y];int ret=size[f[A]];if(f[A]!=f[B])ret+=size[f[B]];if(ret>ans)ans=ret,px=x,py=y; } int main(){scanf("%d%d",&n,&m);for(i=1;i<=n;i++)scanf("%s",a[i]+1);for(i=1;i<=n;i++)for(j=1;j<=m;j++)a[i][j]=a[i][j]=='+';for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(a[i][j]){id[i][j]=++cnt;f[cnt]=cnt;}for(i=1;i<=n;i++)for(j=1;j<=m;j++){left[i][j]=a[i][j]?id[i][j]:left[i][j-1];up[i][j]=a[i][j]?id[i][j]:up[i-1][j];}for(i=n;i;i--)for(j=m;j;j--){right[i][j]=a[i][j]?id[i][j]:right[i][j+1];down[i][j]=a[i][j]?id[i][j]:down[i+1][j];}for(i=1;i<=n;i++){k=0;for(j=1;j<=m;j++)if(a[i][j]){if(k)merge(k,id[i][j]);k=id[i][j];}}for(i=1;i<=m;i++){k=0;for(j=1;j<=n;j++)if(a[j][i]){if(k)merge(k,id[j][i]);k=id[j][i];}}for(i=1;i<=cnt;i++)size[F(i)]++;for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(!a[i][j])gao(i,j);if(ans<=0)puts("0");else printf("%d\n%d %d",ans,px,py); }

　　

轉載于:https://www.cnblogs.com/clrs97/p/7669673.html

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