2013 大连海事大学腾讯马拉松 div2第一场部分题解
1题：给一张正方形的纸，沿中间折叠N次，最后会得到一些小的方格，给出折叠次数和最终得到的最小方格的边长，问原来正方形的面积。其实推出公式用快速幂就解决了。。1001 折纸片
/**********************
* author:crazy_石头
* Pro:
* algorithm:
* Time:0ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
#define LL long long
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29
const int maxn=10000+5;
const LL mod=10000007 ;
inline LL quick_pow(LL a,LL b)
{
LL ret=1;
if(b==0)return 1;
else if(b==1)return a%mod;
else if(b==2)return (a%mod*a)%mod;
while(b)
{
if(b&1)
ret=(LL)ret*a%mod;
b>>=1;
a=(LL)a*a%mod;
}
return ret;
}
int main()
{
int n,k;
while(cin>>n>>k)
{
if(n&1)n=n/2+1;
else n=n/2;
LL res=((k%mod)*(quick_pow(2,n)%mod))%mod;
cout<<res*res%mod<<endl;
}
return 0;
}
2,简单DP，状态转移方程：dp[i]=dp[i-1]+dp[i-2]*2; 1002 升旗
/**********************
* author:crazy_石头
* Pro:
* algorithm:
* Time:0ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <set>
using namespace std;
#define LL long long
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29
const int maxn=600+5;
LL dp[maxn];
int main()
{
int n;
dp[1]=3;
dp[2]=6;
dp[3]=6;
for(int i=4;i<=50;i++)
dp[i]=dp[i-1]+dp[i-2]*2;
while(cin>>n)
cout<<dp[n]<<endl;
return 0;
}
3.给出一个数字，开始代表n盏灭着的灯，从1~n依次按一些灯，每按一个数，它的倍数的灯的状态会跟随它发生相同的变化，问按完后，那些灯还亮着，开始暴力的搞，直接TLE了。。。。现在这个代码是别人的，还不知道为何能这么做。。。1003 按灯
/**********************
* author:crazy_石头
* Pro:
* algorithm:string
* Time:0ms
* Judge Status:Accepted
***********************/
#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
const int mod=10000007;
int main()
{
int n;
while(cin>>n)
{
int t=sqrt(n+0.5);
cout<<t<<endl;
}
return 0;
}
4.一个正方形，上面有一些坏点，问不覆盖坏点的最大正方形区域的面积，好像是这样，直接dp搞。。dp[i][j]=min(dp[i-1][j],dp[i-1][j-1],dp[i][j-1])+1;1004 计算面积
/**********************
* author:crazy_石头
* algorithm:string
* Time:0ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <set>
#include <stack>
#include <string>
using namespace std;
#define LL long long
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29
const int maxn=1000+5;
const int maxm=100+5;
const LL mod=10000007 ;
LL dp[maxn][maxn];
int main()
{
int m,n;
while(~scanf("%d%d",&n,&m))
{
rep(i,1,n)
{
rep(j,1,n)
dp[i][j]=1;
}
rep(i,1,m)
{
int a,b;
cin>>a>>b;
dp[a][b]=0;
}
LL ret=-1;
rep(i,1,n)
{
rep(j,1,n)
{
if(dp[i][j])
{
dp[i][j]=min(dp[i][j-1],min(dp[i-1][j],dp[i-1][j-1]))+1;
ret=max(dp[i][j],ret);
}
}
}
cout<<ret*ret<<endl;
}
return 0;
}
6.搜索？最短路。。。九野巨巨的dfsTLE了啊啊啊啊啊啊，不科学啊啊啊啊1006 飞棋
8.比较复杂，好像是一般图的带花树匹配，果断上bin哥模板啊，我艹，好爽，基本不改。。。1008
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 250;
int N;
bool Graph[MAXN][MAXN];
int Match[MAXN];
bool InQueue[MAXN],InPath[MAXN],InBlossom[MAXN];
int Head,Tail;
int Queue[MAXN];
int Start,Finish;
int NewBase;
int Father[MAXN],Base[MAXN];
int Count;
int m;
void CreateGraph()
{
int u,v;
memset(Graph,false,sizeof(Graph));
while(m--)
{
scanf("%d%d",&u,&v);
Graph[u][v] = Graph[v][u] = true;
}
}
void Push(int u)
{
Queue[Tail] = u;
Tail++;
InQueue[u] = true;
}
int Pop()
{
int res = Queue[Head];
Head++;
return res;
}
int FindCommonAncestor(int u,int v)
{
memset(InPath,false,sizeof(InPath));
while(true)
{
u = Base[u];
InPath[u] = true;
if(u == Start) break;
u = Father[Match[u]];
}
while(true)
{
v = Base[v];
if(InPath[v])break;
v = Father[Match[v]];
}
return v;
}
void ResetTrace(int u)
{
int v;
while(Base[u] != NewBase)
{
v = Match[u];
InBlossom[Base[u]] = InBlossom[Base[v]] = true;
u = Father[v];
if(Base[u] != NewBase) Father[u] = v;
}
}
void BloosomContract(int u,int v)
{
NewBase = FindCommonAncestor(u,v);
memset(InBlossom,false,sizeof(InBlossom));
ResetTrace(u);
ResetTrace(v);
if(Base[u] != NewBase) Father[u] = v;
if(Base[v] != NewBase) Father[v] = u;
for(int tu = 1; tu <= N; tu++)
if(InBlossom[Base[tu]])
{
Base[tu] = NewBase;
if(!InQueue[tu]) Push(tu);
}
}
void FindAugmentingPath()
{
memset(InQueue,false,sizeof(InQueue));
memset(Father,0,sizeof(Father));
for(int i = 1;i <= N;i++)
Base[i] = i;
Head = Tail = 1;
Push(Start);
Finish = 0;
while(Head < Tail)
{
int u = Pop();
for(int v = 1; v <= N; v++)
if(Graph[u][v] && (Base[u] != Base[v]) && (Match[u] != v))
{
if((v == Start) || ((Match[v] > 0) && Father[Match[v]] > 0))
BloosomContract(u,v);
else if(Father[v] == 0)
{
Father[v] = u;
if(Match[v] > 0)
Push(Match[v]);
else
{
Finish = v;
return;
}
}
}
}
}
void AugmentPath()
{
int u,v,w;
u = Finish;
while(u > 0)
{
v = Father[u];
w = Match[v];
Match[v] = u;
Match[u] = v;
u = w;
}
}
void Edmonds()
{
memset(Match,0,sizeof(Match));
for(int u = 1; u <= N; u++)
if(Match[u] == 0)
{
Start = u;
FindAugmentingPath();
if(Finish > 0)AugmentPath();
}
}
void PrintMatch()
{
Count = 0;
for(int u = 1; u <= N;u++)
if(Match[u] > 0)
Count++;
Count/=2;
cout<<Count<<endl;
}
int main()
{
while(cin>>N>>m)
{
CreateGraph();
Edmonds();
PrintMatch();
}
return 0;
}
9.简单的栈应用，逆序输出字符串。。1009 字符游戏
/**********************
* author:crazy_石头
* Pro:
* algorithm:string
* Time:0ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <set>
#include <stack>
#include <string>
using namespace std;
#define LL long long
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29
const int maxn=600000+5;
const int maxm=100+5;
const LL mod=10000007 ;
stack<string> s;
int main()
{
int test,ha=1;
cin>>test;
getchar();
while(test--)
{
char c=getchar();
string t;
t="";
while(c!='\n')
{
if(c!=' ')t+=c;
else
{
s.push(t);
t="";
}
c=getchar();
}
if(t!="")s.push(t);
printf("Case %d:",ha++);
while(!s.empty())
{
printf(" %s",s.top().c_str());
s.pop();
}
puts("");
}
return 0;
}
11.无耻的打表交过。。。。。1011 数独
/**********************
* author:crazy_石头
* Pro:SGU 117
* algorithm:quick pow
* Time:0ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
#define LL long long
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29
const int maxn=10000+5;
const LL mod=10000007 ;
int main()
{
int n;
while(cin>>n)
{
if(n<=2)puts("1");
else if(n==3)puts("2");
else if(n==4)puts("24");
else if(n==5)puts("1344");
else if(n==6)puts("1128960");
else if(n==7)puts("12198297600");
}
return 0;
}
13.一个二分题目，看到最大最小，基本就可以确定是二分了。。。有n个group，向m个箱子里面投票，问怎么投票，可以使得到的最多票的那个盒子票数最小。1013 选举主席
/**********************
* author:crazy_石头
* Pro:
* algorithm:string
* Time:0ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <set>
#include <stack>
#include <string>
using namespace std;
#define LL long long
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29
const int maxn=600000+5;
const int maxm=100+5;
const LL mod=10000007 ;
int n,b,a[maxn];
inline LL cal(LL x) {
LL ret=0;
rep(i,0,n-1)
ret+=(a[i]+x-1)/x;
return ret;
}
inline LL Binary(LL l, LL r)
{
while(l<=r)
{
LL mid=(l+r)>>1;
if(cal(mid)<=b)
r=mid-1;
else
l=mid+1;
}
return l;
}
int main()
{
while(scanf("%d%d",&n,&b)!=EOF&&~n)
{
rep(i,0,n-1)scanf("%d",&a[i]);
printf("%d\n",Binary(0,INF));
}
return 0;
}