HZNU Training 1 for Zhejiang Provincial Collegiate Programming Contest
赛后总结:
TJ:今天我先到实验室,开始看题,一眼就看了一道防AK的题目,还居然觉得自己能做wwww。然后金姐和彭彭来了以后,我和他们讲了点题目。然后金姐开始搞dfs,我和彭彭看榜研究F题。想了很久脑子都炸了,做不下去了。然后在等金姐做完以后,彭彭说了D题题意,然后金姐开始写另外一道搜索题。做完两道搜索以后,金姐让我写那道线段树的题目。我搞了很久。。。然后超时了,因为要用素数筛,然后加了素数筛也做不出呜呜。金姐和彭彭把Ant的题目公式给搞出来了。开始敲。后来是因为取模的问题,就一直没搞成功。比赛吃完饭回来搞了一下,就。。ac了。
呜呜呜,我要更努力,不给队友拖后腿。
JZH:今天又迟到了,上完体育课要从a区到b区,好痛苦。刚开始先看了B题,想到了LCIS,结果n太大不能dp,然后就没思路了。。。。然后彭彭让我去做了C题,但是不够仔细,WA了三发才A。做完之后,看了下F题,没思路,听彭彭讲了D题题意,就去做了D题,结果忘记用优先队列和把队列清空,又WA了两发才A (明明打的时候提醒过自己要用优先队列)。然后就去推H题的公式了,结果推出来又因为取模的原因一直过不了。
真是曲折的一天。继续努力!
PJW:到实验室,题目都还没打开,谭总就来跟我讲了一道防AK题。但也忘了为啥没有死磕。跟了榜,我看了D,让金姐看了C,然后。。。我理解错了题意,觉得D题我来解决可能有点难度,就让金姐敲了。。F题是和谭总一起看的,觉得好绕啊,就是没想到DP(虽然贪心也能做)后来。。我开了一道线段树的题目,谭总开了一道Ant,然而是谭总打得线段树,我和金姐看的Ant(我一边继续磕F题,一边帮金姐验算)第一遍开头就错了。。后来因为取模问题,我改代码的时候有想到要分类考虑。。然而可能还是觉得自己“强”的不用分类,就结束了还是WA。去吃饭的路上,都在讨论为什么WA,然后就得出了分类讨论。
我一直都是个打酱油的~~
题解:
给你一块2*2*2的魔方,问你在n步内能拼成的最多的面有多少。
一共有6*2种旋转方法,但是因为一边旋转,相当于另一边也在旋转,所以只需要列举6种情况。
借用一下https://www.cnblogs.com/linkzijun/p/7050040.html的图。
然后dfs暴力一下就好了。注意数字不要填错wwww。填错了,检查的眼睛痛。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<set> #include<map> #include<string> #include<vector> #include<ctime> #include<stack> using namespace std; const int maxn = 2e5 + 50; #define MAX_DISTANCE 0x3f3f3f3f #define mm(a,b) memset(a,b,sizeof(a)) #define ll long long #define SIGN(A) ((A > 0) ? 1 : -1) #define NO_DISTANCE 1000000 const int INF = 0x3f3f3f3f; #define LL long long int #define mod 1000000007 int gcd(int a, int b) { return a == 0 ? b : gcd(b % a, a); } int maxx; int N; int trans[7][24] = { {0,1,8,14,4,3,7,13,17,9,10,2,6,12,16,15,5,11,18,19,20,21,22,23}, {0,1,11,5,4,16,12,6,2,9,10,17,13,7,3,15,14,8,18,19,20,21,22,23}, {6,1,12,3,5,11,16,7,8,9,4,10,18,13,14,15,20,17,22,19,0,21,2,23}, {20,1,22,3,10,4,0,7,8,9,11,5,2,13,14,15,6,17,12,19,16,21,18,23}, {1,3,0,2,23,22,4,5,6,7,10,11,12,13,14,15,16,17,18,19,20,21,9,8}, {2,0,3,1,6,7,8,9,23,22,10,11,12,13,14,15,16,17,18,19,20,21,5,4}, }; void dfs(int a[],int step) { int num = 0; if (a[0] == a[1] && a[0] == a[2] && a[0] == a[3]) num++; if (a[4] == a[5] && a[4] == a[10] && a[4] == a[11]) num++; if (a[6] == a[7] && a[6] == a[12] && a[6] == a[13]) num++; if (a[8] == a[9] && a[8] == a[14] && a[8] == a[15]) num++; if (a[16] == a[17] && a[16] == a[18] && a[16] == a[19]) num++; if (a[20] == a[21] && a[20] == a[22] && a[20] == a[23]) num++; maxx = max(maxx, num); if (step == N || maxx == 6) return; int m[24]; for (int i = 0; i < 6; i++) { for (int j = 0; j < 24; j++) { m[j] = a[trans[i][j]]; } dfs(m, step + 1); } } int main() { while (scanf("%d", &N) != EOF) { maxx = 0; int a[50]; for (int i = 0; i < 24; i++) cin >> a[i]; dfs(a, 0); cout << maxx << endl; } system("pause"); return 0; }
农民开始播种。在n*m的田地上有几块石头,一开始机器从左上角点开始出发,问是否开机器,经过田地上的每个点只用一次,大石头不能到达。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<set> #include<map> #include<string> #include<vector> #include<ctime> #include<stack> using namespace std; const int MAXN = 2e5+50; #define MAX_DISTANCE 0x3f3f3f3f #define mm(a,b) memset(a,b,sizeof(a)) #define ll long long #define SIGN(A) ((A > 0) ? 1 : -1) #define NO_DISTANCE 1000000 int n, m, sum; char a[100][100]; int vis[100][100]; int dir[4][2] = { 1,0,-1,0,0,1,0,-1 }; bool check(int x, int y) { if (x < 0 || x >= n || y < 0 || y >= m || vis[x][y] || a[x][y] == 'S') return false; return true; } bool dfs(int x, int y, int k) { if (k <= 1) return true; vis[x][y] = 1; for (int i = 0; i < 4; ++i) { int xx = x + dir[i][0]; int yy = y + dir[i][1]; if (check(xx, yy)) { if (dfs(xx, yy, k - 1)) return true; vis[xx][yy] = 0; } } return false; } int main() { while (scanf("%d %d", &n, &m), n || m) { mm(vis, 0); for (int i = 0; i < n; ++i) scanf("%s", a[i]); sum = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { if (a[i][j] == '.') sum++; } } if (dfs(0, 0, sum)) printf("YES\n"); else printf("NO\n"); } //system("pause"); return 0; }
天使的朋友营救天使,经过一个点用花费1时间,打败guard需要花费1(可打败所有guard),问最后营救天使最少需要花费多少时间。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<set> #include<map> #include<string> #include<vector> #include<ctime> #include<stack> using namespace std; const int MAXN = 2e5 + 50; #define MAX_DISTANCE 0x3f3f3f3f #define mm(a,b) memset(a,b,sizeof(a)) #define ll long long #define SIGN(A) ((A > 0) ? 1 : -1) #define NO_DISTANCE 1000000 int n, m, sum; char a[300][300]; int vis[300][300]; int dir[4][2] = { 1,0,-1,0,0,1,0,-1 }; bool check(int x, int y) { if (x < 0 || x >= n || y < 0 || y >= m || vis[x][y] || a[x][y] == '#') return false; return true; } struct node { int x, y, t; bool friend operator <(node a, node b) { return a.t > b.t; } }; priority_queue<node>q; int bfs() { while (!q.empty()) { node now = q.top(); q.pop(); //cout << now.x << " " << now.y <<" "<<now.t<< endl; if (a[now.x][now.y] == 'a') return now.t; for (int i = 0; i < 4; ++i) { int xx = now.x + dir[i][0]; int yy = now.y + dir[i][1]; if (check(xx, yy)) { vis[xx][yy] = 1; if (a[xx][yy] == 'x') q.push(node{ xx,yy,now.t + 2 }); else q.push(node{ xx,yy,now.t + 1 }); } } } return -1; } int main() { while (~scanf("%d %d", &n, &m)) { mm(vis, 0); while (!q.empty()) q.pop(); for (int i = 0; i < n; ++i) scanf("%s", a[i]); sum = 0; for (int i = 0; i < n; ++i) { for (int j = 0; j < m; ++j) { if (a[i][j] == 'r') { q.push(node{ i,j,0 }); vis[i][j] = 1; } } } int time = bfs(); if (time == -1) printf("Poor ANGEL has to stay in the prison all his life.\n"); else printf("%d\n", time); } //system("pause"); return 0; }
F - Cake ZOJ - 3905
A和B去买蛋糕,一共买了n块蛋糕(n为偶数)。每块蛋糕对A和B的价值是不一样的。每次A从n块蛋糕里选择两块,然后B选择一块对于他来说,价值最大的值。求n/2轮以后,A获得的最大的价值为多少。
是一道dp题目。dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]+cake[i].a)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<set> #include<map> #include<string> #include<vector> #include<ctime> #include<stack> using namespace std; const int maxn = 2e5 + 50; #define MAX_DISTANCE 0x3f3f3f3f #define mm(a,b) memset(a,b,sizeof(a)) #define ll long long #define SIGN(A) ((A > 0) ? 1 : -1) #define NO_DISTANCE 1000000 const int INF = 0x3f3f3f3f; #define LL long long int #define mod 1000000007 int gcd(int a, int b) { return a == 0 ? b : gcd(b % a, a); } const int N = 10; struct Cake { int a; int b; }cake[1000]; int dp[1000][1000]; bool cmp(Cake aa, Cake bb) { return aa.b > bb.b; } int main() { int T; cin >> T; while (T--) { memset(dp, 0, sizeof(dp)); int n; cin >> n; for (int i = 1; i <= n; i++) cin >> cake[i].a >> cake[i].b; sort(cake + 1, cake + 1 + n, cmp); for (int i = 1; i <= n; i++) { for (int j = 0; j <= i / 2; j++) { if (j > 0) dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + cake[i].a); else dp[i][j] = dp[i - 1][j]; } } cout << dp[n][n / 2] << endl; } //system("pause"); return 0; }
G - Prime Query ZOJ - 3911
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#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #define mid(a,b) ((a+b)>>1) #define LL int #define maxn 110000 #define IN freopen("in.txt","r",stdin); using namespace std; char is_prime[maxn*100]; void sieve() { int m=(int)sqrt((maxn*100)+0.5); fill(is_prime,is_prime+(maxn*100),1); is_prime[0]=is_prime[1]=0; for(int i=2;i<=m;i++) if(is_prime[i]) for(int j=i*i;j<(maxn*100);j+=i) is_prime[j]=0; } int n,q; LL num[maxn]; struct Tree { int left,right; LL sum; LL val,lazy; }tree[maxn<<2]; /*递归建树*/ void build(int i,int left,int right) { tree[i].left=left; tree[i].right=right; tree[i].lazy=0; if(left==right){ tree[i].val=num[left]; tree[i].sum=(is_prime[num[left]]? 1:0); return ; } int mid=mid(left,right); build(i<<1,left,mid); build(i<<1|1,mid+1,right); tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum; } /*区间修改,标记下传:每当访问到当前结点的子节点时,下传标记*/ void pushdown(int i) { if(tree[i].lazy){ int tmp = (is_prime[tree[i].lazy]? 1:0); tree[i<<1].val=tree[i].lazy; tree[i<<1|1].val=tree[i].lazy; tree[i<<1].lazy=tree[i].lazy; tree[i<<1|1].lazy=tree[i].lazy; tree[i<<1].sum=(tree[i<<1].right-tree[i<<1].left+1)*tmp; tree[i<<1|1].sum=(tree[i<<1|1].right-tree[i<<1|1].left+1)*tmp; tree[i].lazy=0; /*下传后清零*/ } } /*区间修改,d为改变量*/ void update(int i,int left,int right,LL d) { if(tree[i].left==left&&tree[i].right==right) { int tmp = (is_prime[d]? 1:0); tree[i].sum=(right-left+1)*tmp; tree[i].val=d; tree[i].lazy=d; return ; } pushdown(i); int mid=mid(tree[i].left,tree[i].right); if(right<=mid) update(i<<1,left,right,d); else if(left>mid) update(i<<1|1,left,right,d); else { update(i<<1,left,mid,d); update(i<<1|1,mid+1,right,d); } tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum; } /*单点修改,d为改变量*/ void update(int i,int x,LL d) { if(tree[i].left==tree[i].right){ tree[i].val+=d; tree[i].sum=(is_prime[tree[i].val]? 1:0); return; } pushdown(i); int mid=mid(tree[i].left,tree[i].right); if(x<=mid) update(i<<1,x,d); else update(i<<1|1,x,d); tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum; } /*区间结果查询*/ LL query(int i,int left,int right) { if(tree[i].left==left&&tree[i].right==right) return tree[i].sum; pushdown(i); int mid=mid(tree[i].left,tree[i].right); if(right<=mid) return query(i<<1,left,right); else if(left>mid) return query(i<<1|1,left,right); else return query(i<<1,left,mid)+query(i<<1|1,mid+1,right); } int main(int argc, char const *argv[]) { //IN; sieve(); int t;scanf("%d",&t); while(t--) { scanf("%d %d",&n,&q); for(int i=1;i<=n;i++) scanf("%d",&num[i]); build(1,1,n); while(q--) { char c; while(c=getchar()){ if(c=='A'||c=='R'||c=='Q') break; } if(c=='A'){ int v,l; scanf("%d %d",&v,&l); update(1,l,v); } if(c=='R'){ int a,l,r; scanf("%d %d %d",&a,&l,&r); update(1,l,r,a); } if(c=='Q'){ int l,r; scanf("%d %d",&l,&r); printf("%d\n", query(1,l,r)); } } } return 0; }
小蚂蚁从立方体角上的一点爬到立方体对面的点,给你一条最长的一边n长度,另外两条边1<=a,b<=n,问所有最短长度L^2的合(答案对1e9+7取模)。
公式:
L^2=(n*n)*(n*(n+1))/2+(n*n)*(n+1)*(n+1)/2+n*(n+1)*(2*n+1)*(n+2)/6;
(具体看代码)
因为n的范围在1~10^14,所以计算取模的时候一定要小心。还有在/6的时候需要进行特判。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<set> #include<map> #include<string> #include<vector> #include<ctime> #include<stack> using namespace std; const int maxn = 2e5 + 50; #define MAX_DISTANCE 0x3f3f3f3f #define mm(a,b) memset(a,b,sizeof(a)) #define ll long long #define SIGN(A) ((A > 0) ? 1 : -1) #define NO_DISTANCE 1000000 const int INF = 0x3f3f3f3f; #define LL long long int #define mod 1000000007 int main() { int t; ll n; cin >> t; while (t--) { scanf("%lld", &n); n %= mod; ll ans = (n*(n + 1) / 2) % mod*((n * n) % mod); ans = (ans + (((n*(n + 1) / 2) % mod)*((n*(n + 1) / 2) % mod)) % mod) % mod; if (n % 3 == 0||(n+1)%3==0) { ans = (ans + (((n*(n + 1) / 6) % mod)*(((2 * n + 1)*(n + 2)) % mod))) % mod; } else { ans = (ans + (((n*(n + 1) / 2) % mod)*(((2 * n + 1)*(n + 2)/3) % mod))) % mod; } //ans = (ans + (((n*(n + 1) / 2) % mod)*(((2 * n + 1)*(n + 2)) % mod) / 3)) % mod; // // ll ans = ((((n * n) % mod )*n % mod) *(n + 1)) % mod / 2; //ans = (ans + ((((n * n % mod) *(n + 1) % mod)*(n + 1) % mod)%mod )/ 4) % mod; //ans = (ans + (((n * (n + 1) % mod)*((2 * n + 1) % mod ) % mod)%mod*(n + 2) % mod) / 6) % mod; cout << ans << endl; } //system("pause"); return 0; }
K - Asteroids POJ - 3041
一个n*n的二维坐标上,有k个星星,给你星星的位置(x,y都为整数),可以一整排一整列消灭星星,问最少需要几步。
思路:https://blog.csdn.net/u013480600/article/details/38615197
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<set> #include<map> #include<string> #include<vector> #include<ctime> #include<stack> using namespace std; const int maxn = 1000; #define MAX_DISTANCE 0x3f3f3f3f #define mm(a,b) memset(a,b,sizeof(a)) #define ll long long #define SIGN(A) ((A > 0) ? 1 : -1) #define NO_DISTANCE 1000000 const int INF = 0x3f3f3f3f; #define LL long long int #define mod 1000000007 int gcd(int a, int b) { return a == 0 ? b : gcd(b % a, a); } struct Max_Match { int n; bool g[maxn][maxn]; bool vis[maxn*2]; int left[maxn*2]; void init(int n) { this->n = n; memset(g, 0, sizeof(g)); memset(left, -1, sizeof(left)); } bool match(int u) { for (int v = 1; v <= n; v++) { if (g[u][v] && !vis[v]) { vis[v] = true; if (left[v] == -1 || match(left[v])) { left[v] = u; return true; } } } return false; } int solve() { int ans = 0; for (int i = 1; i <= n; i++) { memset(vis, 0, sizeof(vis)); if (match(i)) ans++; } return ans; } }MM; int main() { int n, k; scanf("%d %d", &n, &k); MM.init(n); while (k--) { int u, v; scanf("%d %d", &u, &v); MM.g[u][v] = true; } printf("%d\n", MM.solve()); return 0; }