2017 Multi-University Training Contest

题目链接：http://acm.hdu.edu.cn/contests/contest_show.php?cid=759

Add More Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2245    Accepted Submission(s): 1053

Problem Description

There is a youngster known for amateur propositions concerning several mathematical hard problems.

Nowadays, he is preparing a thought-provoking problem on a specific type of supercomputer which has ability to support calculations of integers between 0 and (2m−1) (inclusive).

As a young man born with ten fingers, he loves the powers of 10 so much, which results in his eccentricity that he always ranges integers he would like to use from 1to 10k (inclusive).

For the sake of processing, all integers he would use possibly in this interesting problem ought to be as computable as this supercomputer could.

Given the positive integer m, your task is to determine maximum possible integer k that is suitable for the specific supercomputer.

 

Input

The input contains multiple test cases. Each test case in one line contains only one positive integer m, satisfying 1≤m≤105.

 

Output

For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

 

Sample Input

164

 

Sample Output

Case #1: 0Case #2: 19

 

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解析：等式两边同时取模，转换下就可以了

代码：

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1000009;const int mod = 1e9+7ll;int main(){    int m, cnt = 0;    while(~scanf("%d", &m))    {        int ans = m * log10(2.0);        printf("Case #%d: %d\n", ++cnt, ans);    }    return 0;}

Balala Power!

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4809    Accepted Submission(s): 387

Problem Description

Talented Mr.Tang has n strings consisting of only lower case characters. He wants to charge them with Balala Power (he could change each character ranged from a to z into each number ranged from 0 to 25, but each two different characters should not be changed into the same number) so that he could calculate the sum of these strings as integers in base 26 hilariously.

Mr.Tang wants you to maximize the summation. Notice that no string in this problem could have leading zeros except for string "0". It is guaranteed that at least one character does not appear at the beginning of any string.

The summation may be quite large, so you should output it in modulo 109+7.

 

Input

The input contains multiple test cases.

For each test case, the first line contains one positive integers n, the number of strings. (1≤n≤100000)

Each of the next n lines contains a string si consisting of only lower case letters. (1≤|si|≤100000,∑|si|≤106)

 

Output

For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

 

Sample Input

1a2aabb3abaabc

 

Sample Output

Case #1: 25Case #2: 1323Case #3: 18221

 

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解析：每个字符对答案的贡献都可以看作一个 26 进制的数字，问题相当于要给这些贡献加一个 0 到 25 的权重使得答案最大。最大的数匹配 25，次大的数匹配 24，依次类推。排序后这样依次贪心即可，唯一注意的是不能出现前导 0。

代码：

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1000009;const LL mod = 1e9+7ll;//typedef pair<int, char> P;int a[100100][33];char s[100100];int x[33], vis[33], m;bool cmp(int p, int q){    for(int j = m-1; j >= 0; j--)    {        if(a[j][p] < a[j][q]) return false;        else if(a[j][p] > a[j][q]) return true;    }    return true;}void Sort(){    for(int i = 0; i < 26; i++) x[i] = i;    for(int i = 0; i < 25; i++)    {        for(int j = i + 1; j < 26; j++)        {            if(cmp(x[i], x[j]))            {                int t = x[i];                x[i] = x[j];                x[j] = t;            }        }    }}int main(){    int n, cnt = 0;    while(~scanf("%d", &n))    {        m = 0;        memset(vis, 0, sizeof(vis));        memset(a, 0, sizeof(a));        for(int i = 1; i <= n; i++)        {            scanf(" %s", s);            int len = strlen(s);            m = max(m, len);            int k = 0;            vis[s[0]-'a'] = 1;            for(int j = len - 1; j >= 0; j--){a[k][s[j] - 'a']++;k++;}        }        int t[33];        memset(t, 0, sizeof(t));        for(int i = 0; i < m-1; i++)        {            for(int j = 0; j < 26; j++)            {                int cur = (t[j] + a[i][j]);                a[i][j] = cur % 26;                t[j] = cur / 26;            }        }        for(int i = 0; i < 26; i++) a[m-1][i] += t[i];        Sort();        LL ans = 0ll;        int k = 1, p;        for(int i = 0; i < 26; i++)        {            int j = x[i];            if(vis[j] == 0)            {                p = j;                break;            }        }        for(int i = 0; i < 26; i++)        {            int cur = x[i];            if(p == cur) continue;            LL num = 0ll;            for(int j = m - 1; j >= 0; j--) num = (num * 26 + k * a[j][cur]) % mod;            ans = (ans + num) % mod;            k++;        }        printf("Case #%d: %lld\n", ++cnt, ((ans%mod)+mod)%mod);    }    return 0;}

KazaQ's Socks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2375    Accepted Submission(s): 1074

Problem Description

KazaQ wears socks everyday.

At the beginning, he has n pairs of socks numbered from 1 to n in his closets. 

Every morning, he puts on a pair of socks which has the smallest number in the closets. 

Every evening, he puts this pair of socks in the basket. If there are n−1 pairs of socks in the basket now, lazy KazaQ has to wash them. These socks will be put in the closets again in tomorrow evening.

KazaQ would like to know which pair of socks he should wear on the k-th day.

 

Input

The input consists of multiple test cases. (about 2000)

For each case, there is a line contains two numbers n,k (2≤n≤109,1≤k≤1018).

 

Output

For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

 

Sample Input

3 73 64 9

 

Sample Output

Case #1: 3Case #2: 1Case #3: 2

 

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解析：找下规律即可，比如n = 4, 123 412312 412312 412312  ，注意特判n=2

代码：

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1000009;const int mod = 1e9+7ll;LL solve(LL n, LL k){    if(n == 2&&k&1) return 1;    else if(n == 2) return 2;    if(k <= n - 1) return k;    k -= n - 1;    k %= 2*(n-1);    if(k == 0) return n - 2;    if(k <= n - 1)    {        if(k == 1) return n;        return k - 1;    }    k -= n - 1;    if(k == 1) return n-1;    return k - 1;}int main(){    LL n, k;    int cnt = 0;    while(~scanf("%lld%lld", &n, &k))    {        printf("Case #%d: ", ++cnt);        cout << solve(n, k) << endl;    }    return 0;}