2014 Multi-University Training Contest 7 1003 1005

http://acm.hdu.edu.cn/showproblem.php?pid=4937

Lucky Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1123    Accepted Submission(s): 333

Problem Description

“Ladies and Gentlemen, It’s show time! ”

“A thief is a creative artist who takes his prey in style... But a detective is nothing more than a critic, who follows our footsteps...”

Love_Kid is crazy about Kaito Kid , he think 3(because 3 is the sum of 1 and 2), 4, 5, 6 are his lucky numbers and all others are not. 

Now he finds out a way that he can represent a number through decimal representation in another numeral system to get a number only contain 3, 4, 5, 6. 

For example, given a number 19, you can represent it as 34 with base 5, so we can call 5 is a lucky base for number 19. 

Now he will give you a long number n(1<=n<=1e12), please help him to find out how many lucky bases for that number. 

If there are infinite such base, just print out -1.

 

Input

There are multiply test cases.

The first line contains an integer T(T<=200), indicates the number of cases. 

For every test case, there is a number n indicates the number.

 

Output

For each test case, output “Case #k: ”first, k is the case number, from 1 to T , then, output a line with one integer, the answer to the query.

 

Sample Input

21019

 

Sample Output

Case #1: 0Case #2: 1
Hint
10 shown in hexadecimal number system is another letter different from ‘0’-‘9’, we can represent it as ‘A’, and you can extend to other cases.
 

 

Author

UESTC

 

Source

2014 Multi-University Training Contest 7

 

给一个数N,求N能否转化成一个n进制数X，且X只由3,4,5,6组成。求X的个数。详见代码注释。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;ll MIN;//MIN代表最小进制数，比如如果N能表示成两位的X进制数，那么当N表示成三位的X’进制时，X’<X;所以记录下最小的X，为之后进制数高时减少复杂度//n能表示成两位数时bool cal2(ll n,int i,int j){    ll k=n-j;    ll base=k/i;    if(k%i!=0) return false;    else {        if(base>i&&base>j) {MIN=min(MIN,base);return true;}    }    return false;}//n能表示成三位数时bool cal3(ll n,int i,int j,int k){    ll kk=k-n;    ll deta=j*j-4*i*kk;    if(deta<0) return false;    ll dd=sqrt(deta);    if(dd*dd!=deta) return false;    ll xx=(-j-dd)/(2*i);    if((-j-dd)%(2*i)==0&&xx>i&&xx>j&&xx>k){        MIN=min(xx,MIN);        return true;    }    xx=(-j+dd)/(2*i);    if((-j+dd)%(2*i)==0&&xx>i&&xx>j&&xx>k){        MIN=min(xx,MIN);        return true;    }    return false;}bool cal4(ll n,ll i){    ll k=n%i;    while(n){        if(k!=3&&k!=4&&k!=5&&k!=6) return false;        n/=i;        k=n%i;    }    return true;}int main(){    //freopen("1003.in","r",stdin);    //freopen("10032.out","w",stdout);    int t;    scanf("%d",&t);    for(int ii=1;ii<=t;++ii){        ll n;        scanf("%I64d",&n);        MIN=n;        printf("Case #%d: ",ii);        if(n==3||n==4||n==5||n==6){            printf("-1\n");            continue;        }        int ans=0;        for(int i=3;i<=6;i++)            for(int j=3;j<=6;j++)                if(cal2(n,i,j)) ans++;        for(int i=3;i<=6;i++)            for(int j=3;j<=6;j++)                for(int k=3;k<=6;k++)                    if(cal3(n,i,j,k)) ans++;        //如果两位和3位都没有，那最大的进制就是6*base^(X-1)=10^12,因为之前X=3时不满足，所以此时X最小=4,故max{base}=10^4        for(ll i=4;i<min(MIN,7000ll);i++)            if(cal4(n,i)) ans++;        printf("%d\n",ans);    }    return 0;}

http://acm.hdu.edu.cn/showproblem.php?pid=4939

Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1315    Accepted Submission(s): 386

Problem Description

FSF is addicted to a stupid tower defense game. The goal of tower defense games is to try to stop enemies from crossing a map by building traps to slow them down and towers which shoot at them as they pass.

The map is a line, which has n unit length. We can build only one tower on each unit length. The enemy takes t seconds on each unit length. And there are 3 kinds of tower in this game: The red tower, the green tower and the blue tower. 

The red tower damage on the enemy x points per second when he passes through the tower.

The green tower damage on the enemy y points per second after he passes through the tower.

The blue tower let the enemy go slower than before (that is, the enemy takes more z second to pass an unit length, also, after he passes through the tower.)

Of course, if you are already pass through m green towers, you should have got m*y damage per second. The same, if you are already pass through k blue towers, the enemy should have took t + k*z seconds every unit length.

FSF now wants to know the maximum damage the enemy can get.

 

Input

There are multiply test cases.

The first line contains an integer T (T<=100), indicates the number of cases. 

Each test only contain 5 integers n, x, y, z, t (2<=n<=1500,0<=x, y, z<=60000,1<=t<=3)

 

Output

For each case, you should output "Case #C: " first, where C indicates the case number and counts from 1. Then output the answer. For each test only one line which have one integer, the answer to this question.

 

Sample Input

12 4 3 2 1

 

Sample Output

Case #1: 12
Hint
For the first sample, the first tower is blue tower, and the second is red tower. So, the total damage is 4*(1+2)=12 damage points.
 

 

Author

UESTC

 

Source

2014 Multi-University Training Contest 7

 

dp,当时没有想清楚，要有红色一定放最后是没问题的，接下来就是方程，dp[j][k]表示前j个长度内有k个蓝塔造成的伤害的最大值。前面wa了几次，居然忘记乘法给溢出了，最近特别不细心，静心啊。

dp[j][k]=max(dp[j-1][k]+(ll)(j-k-1)*(t+k*z)*y,dp[j-1][k-1]+(ll)(j-k)*(t+(k-1)*z)*y);

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#define MAXN 1550using namespace std;typedef long long ll;ll dp[MAXN][MAXN];int main(){    int T;    scanf("%d",&T);    memset(dp,0,sizeof(dp));    for(int i = 1; i <= T; ++i) {        int n, x, y, z, t;        scanf("%d%d%d%d%d",&n, &x, &y, &z, &t);        printf("Case #%d: ",i);        ll ans=(ll)n*x*t;        ll damage=0;        for(int j=1;j<=n;j++){            dp[j][0]=dp[j-1][0]+y*(j-1)*t;            ans=max(dp[j][0]+(ll)(n-j)*t*x+(ll)(n-j)*t*y*j,ans);        }        for(int j = 1; j <= n; j++)            for(int k = 1; k <= j; k++){                if(k==j)dp[j][k]=0;                else dp[j][k]=max(dp[j-1][k]+(ll)(j-k-1)*(t+k*z)*y,dp[j-1][k-1]+(ll)(j-k)*(t+(k-1)*z)*y);                damage=dp[j][k]+(ll)(n-j)*(t+k*z)*x+(ll)(n-j)*(t+k*z)*y*(j-k);                ans=max(ans,damage);            }        printf("%I64d\n",ans);        memset(dp,0,sizeof(dp));    }    return 0;}