2017ACM/ICPC广西邀请赛

A Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 765    Accepted Submission(s): 318

Problem Description

You are given a positive integer n, please count how many positive integers k satisfy k^k \leq n.

 

Input

There are no more than 50 test cases.

Each case only contains a positivse integer n in a line.

1 \leq n \leq 10^{18}

 

Output

For each test case, output an integer indicates the number of positive integers k satisfy k^k \leq n in a line.

 

Sample Input

14

 

Sample Output

12

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

Recommend

liuyiding

很简单的一道题，注意ll的数据溢出，可以先打个表，16数据就会溢出，所有只要判断到15

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define MOD 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  vector<int> vi;typedef  long long ll;typedef  unsigned long long  ull;inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};const int N = 3005;ll n;ll k;int main(){    ios_base::sync_with_stdio(false);//    freopen("data.txt","r",stdin);    while(cin >> n)    {        ll ct =0;        for(int i=1;i<=15;i++)        {            ll t = qpow(i,i);//            cout<<t<<" "<<i<<endl;            if(t>n)            {                break;            }            ct++;        }        cout <<ct<<endl;    }    return 0;}

Covering

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 439    Accepted Submission(s): 214

Problem Description

Bob's school has a big playground, boys and girls always play games here after school.

To protect boys and girls from getting hurt when playing happily on the playground, rich boy Bob decided to cover the playground using his carpets.

Meanwhile, Bob is a mean boy, so he acquired that his carpets can not overlap one cell twice or more.

He has infinite carpets with sizes of 1×2 and 2×1, and the size of the playground is 4×n.

Can you tell Bob the total number of schemes where the carpets can cover the playground completely without overlapping?

 

Input

There are no more than 5000 test cases. 

Each test case only contains one positive integer n in a line.

1≤n≤1018

 

Output

For each test cases, output the answer mod 1000000007 in a line.

 

Sample Input

12

 

Sample Output

15

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

Recommend

liuyiding

 

这个是队友想出来的，先推出一个公式，然后用矩阵快速幂直接过

#include <iostream>#include <cstring>#include <cstdio>using namespace std;#define LL long longconst int mod=1000000007;struct matrix{LL x[4][4];};matrix mutimatrix(matrix a,matrix b){matrix temp;memset(temp.x,0,sizeof(temp.x));for(int i=0;i<4;i++)for(int j=0;j<4;j++)for(int k=0;k<4;k++){temp.x[i][j]+=a.x[i][k]*b.x[k][j];temp.x[i][j]%=mod;}return temp;}matrix k_powmatrix(matrix a,LL n){matrix temp;memset(temp.x,0,sizeof(temp.x));for(int i=0;i<4;i++)temp.x[i][i]=1;while(n){if(n&1)temp=mutimatrix(temp,a);a=mutimatrix(a,a);n>>=1;}return temp;}int main(){int T;LL n;while(scanf("%lld",&n)!=EOF){if(n==1){printf("1\n");continue;}if(n==2){printf("5\n");continue;}if(n==3){printf("11\n");continue;}if(n==4){printf("36\n");continue;}matrix st;memset(st.x,0,sizeof(st.x));st.x[0][0]=1;st.x[1][0]=5;st.x[2][0]=1;st.x[3][0]=-1;st.x[0][1]=1;st.x[1][2]=1;st.x[2][3]=1;matrix init;memset(init.x,0,sizeof(init.x));init.x[0][0]=36;init.x[0][1]=11;init.x[0][2]=5;init.x[0][3]=1;st=k_powmatrix(st,n-4);st=mutimatrix(init,st);//for(int i=0;i<4;i++)printf("%lld\n",(st.x[0][0]+mod)%mod);}return 0;}/*    |f[n] f[n-1]|*|a 1|=|f[n+1] f[n]|                  |b 0|*/

CS Course

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 547    Accepted Submission(s): 267

Problem Description

Little A has come to college and majored in Computer and Science.

Today he has learned bit-operations in Algorithm Lessons, and he got a problem as homework.

Here is the problem:

You are giving n non-negative integers a1,a2,⋯,an, and some queries.

A query only contains a positive integer p, which means you 
are asked to answer the result of bit-operations (and, or, xor) of all the integers except ap.

 

Input

There are no more than 15 test cases. 

Each test case begins with two positive integers n and p
in a line, indicate the number of positive integers and the number of queries.

2≤n,q≤105

Then n non-negative integers a1,a2,⋯,an follows in a line, 0≤ai≤109 for each i in range[1,n].

After that there are q positive integers p1,p2,⋯,pqin q lines, 1≤pi≤n for each i in range[1,q].

 

Output

For each query p, output three non-negative integers indicates the result of bit-operations(and, or, xor) of all non-negative integers except ap in a line.

 

Sample Input

3 31 1 1123

 

Sample Output

1 1 01 1 01 1 0

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

Recommend

liuyiding

 

利用位运算的特性，先统计一些所有数的相同位的和，然后再进行判断

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define MOD 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  vector<int> vi;typedef  long long ll;typedef  unsigned long long  ull;inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};const int N = 100005;int sum[35];int n,q;bool dig[N][35];int a[N];int t;int ans[N][3];int vis[N];ll er[35];//int ans[N][3];int main(){//    ios_base::sync_with_stdio(false);//    freopen("data.txt","r",stdin);    er[0] = 1;    for(int i=1;i<35;i++)    {        er[i] = 2*er[i-1];    }    while(~scanf("%d%d",&n,&q))    {        memset(sum,0,sizeof(sum));        memset(ans,0,sizeof(ans));        memset(dig,0,sizeof(dig));        memset(vis,0,sizeof(vis));        memset(a,0,sizeof(a));        int max_dig = 0;        for(int i=0;i<n;i++)        {            int ind=0;            scanf("%d",&t);            a[i] = t;            while(t)            {                dig[i][ind++] = t%2;                t/=2;            }            max_dig = max(max_dig,ind);        }        for(int i=0;i<n;i++)        {            for(int j = 0 ;j < max_dig;j++)            {                sum[j] += dig[i][j];            }        }        int ans1,ans2,ans3;        for(int i = 0;i<q;i++)        {            ans1 = ans2 = ans3 = 0;            scanf("%d",&t);            t--;            if(vis[t])            {                printf("%d %d %d\n",ans[t][0],ans[t][1],ans[t][2]);                continue;            }            vis[t] = 1;            for(int j = 0;j<max_dig;j++)            {                if(dig[t][j]==0&&sum[j]==n-1)                {                    ans[t][0] += er[j];                }                if(dig[t][j]==1&&sum[j]==n)                {                    ans[t][0] += er[j];                }            }            for(int j = 0;j<max_dig;j++)            {                if(dig[t][j]==0&&sum[j])                {                    ans[t][1] += er[j];                }                if(dig[t][j]==1&&sum[j]>1)                {                    ans[t][1] += er[j];                }            }            for(int j = 0;j<max_dig;j++)            {                if(dig[t][j]==0&&sum[j]%2==1)                {                    ans[t][2] += er[j];                }                if(dig[t][j]==1&&(sum[j]-1)%2==1)                {                    ans[t][2] += er[j];                }            }            printf("%d %d %d\n",ans[t][0],ans[t][1],ans[t][2]);        }    }    return 0;}

Duizi and Shunzi

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 538    Accepted Submission(s): 260

Problem Description

Nike likes playing cards and makes a problem of it.

Now give you n integers, ai(1≤i≤n)

We define two identical numbers (eg: 2,2) a Duizi,
and three consecutive positive integers (eg: 2,3,4) a Shunzi.

Now you want to use these integers to form Shunzi and Duizi as many as possible.

Let s be the total number of the Shunzi and the Duizi you formed.

Try to calculate max(s).

Each number can be used only once.

 

Input

The input contains several test cases.

For each test case, the first line contains one integer n(1≤n≤106). 
Then the next line contains n space-separated integers ai (1≤ai≤n)

 

Output

For each test case, output the answer in a line.

 

Sample Input

71 2 3 4 5 6 791 1 1 2 2 2 3 3 362 2 3 3 3 3 61 2 3 3 4 5

 

Sample Output

2432
Hint
Case 1（1，2，3）（4，5，6）Case 2（1，2，3）（1，1）（2，2）（3，3）Case 3（2，2）（3，3）（3，3）Case 4（1，2，3）（3，4，5）
 

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

 

Recommend

liuyiding

 

贪心，先把对子给算减去，奇数留1个，偶数留两个，然后考虑下面这几种情况

1 2 3 3 4 5

1 1 2 2 3 3

1 2 2 3 3

然后遍历一遍就可求解了。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <iomanip>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(int i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)const int inf_int = 2e9;const long long inf_ll = 2e18;#define inf_add 0x3f3f3f3f#define MOD 1000000007#define pb push_back#define mp make_pair#define fi first#define se second#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=5e2+10;using namespace std;typedef  vector<int> vi;typedef  long long ll;typedef  unsigned long long  ull;inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,rx=getchar();return ra*fh;}//#pragma comment(linker, "/STACK:102400000,102400000")ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};const int N = 1000005;int n;int a[N];int num[N];int main(){//    ios_base::sync_with_stdio(false);//    freopen("data.txt","r",stdin);//    freopen("data.txt","w",stdout);    while(scanf("%d",&n)!=EOF)    {        ll ct = 0;        memset(num,0,sizeof(num));        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);            num[a[i]]++;        }        for(int i=1;i<=n;i++)        {            if(num[i]==0) continue;            if(num[i]%2)            {                ct += num[i]/2;                num[i]=1;            }            else            {                ct += (num[i]-2)/2;                num[i]=2;            }        }        for(int i=1;i<=n;i++)        {            if(num[i]==2)            {                ct++;                num[i]=0;            }            else            {                if(num[i]==1&&num[i+1]==1&&num[i+2])                {                    ct++;                    num[i]--;                    num[i+1]--;                    num[i+2]--;                }            }        }        printf("%I64d\n",ct);    }    return 0;}