企鹅很忙系列~（可惜只会做3道题T_T）
A – A
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces 289A
Description
Little penguin Polo adores integer segments, that is, pairs of integers [l; r] (l ≤ r).
He has a set that consists of n integer segments: [l1; r1], [l2; r2], …, [ln; rn]. We know that no two segments of this set intersect. In one move Polo can either widen any segment of the set 1 unit to the left or 1 unit to the right, that is transform [l; r] to either segment[l - 1; r], or to segment [l; r + 1].
The value of a set of segments that consists of n segments [l1; r1], [l2; r2], …, [ln; rn] is the number of integers x, such that there is integer j, for which the following inequality holds, lj ≤ x ≤ rj.
Find the minimum number of moves needed to make the value of the set of Polo’s segments divisible by k.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 105). Each of the following n lines contain a segment as a pair of integers li andri ( - 105 ≤ li ≤ ri ≤ 105), separated by a space.
It is guaranteed that no two segments intersect. In other words, for any two integers i, j (1 ≤ i < j ≤ n) the following inequality holds,min(ri, rj) < max(li, lj).
Output
In a single line print a single integer — the answer to the problem.
Sample Input
Input
2 3
1 2
3 4
Output
2
Input
3 7
1 2
3 3
4 7
Output
0
这个题我看了好久才明白意思啊，英语真拙计，核心意思就是先给出一个数K，再给出n个区间，使得这n个区间的整数的个数和能整除k，如果不能，就要做改变，每一次改变只能增加一个整数，回答最少需要做改变的次数。
#include<stdio.h>
int main()
{
int sum = ,n,k,a,b,i,j;
scanf("%d%d",&n,&k);
while(n--)
{
scanf("%d%d",&a,&b);
sum += b - a + ;
}
if(sum % k == )printf("0\n");
else
{
printf("%d\n",k - sum %k);//k减去k的余数就是这个数还要加几才能达到k，该题中就是要改变的次数
}
return ;
}
B – B
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces 289B
Description
Little penguin Polo has an n × m matrix, consisting of integers. Let’s index the matrix rows from 1 to n from top to bottom and let’s index the columns from 1 to m from left to right. Let’s represent the matrix element on the intersection of row i and column j as aij.
In one move the penguin can add or subtract number d from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so.
Input
The first line contains three integers n, m and d (1 ≤ n, m ≤ 100, 1 ≤ d ≤ 104) — the matrix sizes and the d parameter. Next n lines contain the matrix: the j-th integer in the i-th row is the matrix element aij (1 ≤ aij ≤ 104).
Output
In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print “-1” (without the quotes).
Sample Input
Input
2 2 2
2 4
6 8
Output
4
Input
1 2 7
6 7
Output
-1
给一个M*N的矩阵及一个数D，小企鹅喜欢矩阵，但是喜欢每个数都一样，不过它每次都只能把矩阵中的数加或者减D，输出最少的改变次数，使矩阵的数变成一样的。如果不能达到目标就输出-1，否则输出最少的次数。
用一维数组存矩阵好了（有时候不要被题目欺骗了，一维数组就能做的时候不要只想着开二维数组），然后把数组排序，中间的数就是所有的数希望变成的数（这样改变的次数才能最小）。然后计算次数。最后要说一下怎么判断这个数组不能达到目标呢。因为我写的代码是在存入数组的时候就要判断是否能达到目标，这个时候还不知道所有的数希望变成的数是什么。不如设所有的数希望变成的数是m好了，则a1 + n1 * d = m，a2 + n2 * d = m，两式相减可得（a1 - a2） / d = n2 - n1由此可以得到矩阵中任两个数的差一定是d的倍数，如果存在不是这样的数，则这个矩阵就不符合要求。
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int a[];
int main()
{ int m,n,d,i,x,y = ,sum = ;
cin >> m>> n>> d;
for(i = ;i < n*m;i++)
{
cin >> a[i];
if(i == )continue;
else if((a[i] - a[i - ]) % d != && y==)
{
y = ;
}
}
if(y == )cout << -<<endl;
else
{
sort(a,a+n*m);
int mid = n*m /;
for(i = ;i < n*m;i++)
{
x = abs(a[mid] - a[i]);
sum += x/d;
}
cout << sum<<endl;
}
return ;
}
C – C
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit Status Practice CodeForces 289C
Description
Little penguin Polo adores strings. But most of all he adores strings of length n.
One day he wanted to find a string that meets the following conditions:
- The string consists of n lowercase English letters (that is, the string’s length equals n), exactly k of these letters are distinct.
- No two neighbouring letters of a string coincide; that is, if we represent a string as s = s1s2… sn, then the following inequality holds, si ≠ si + 1(1 ≤ i < n).
- Among all strings that meet points 1 and 2, the required string is lexicographically smallest.
Help him find such string or state that such string doesn’t exist.
String x = x1x2… xp is lexicographically less than string y = y1y2… yq, if either p < q and x1 = y1, x2 = y2, … , xp = yp, or there is such number r(r < p, r < q), that x1 = y1, x2 = y2, … , xr = yr and xr + 1 < yr + 1. The characters of the strings are compared by their ASCII codes.
Input
A single line contains two positive integers n and k(1 ≤ n ≤ 106, 1 ≤ k ≤ 26) — the string’s length and the number of distinct letters.
Output
In a single line print the required string. If there isn’t such string, print “-1” (without the quotes).
Sample Input
Input
7 4
Output
ababacd
Input
4 7
Output
-1
小企鹅喜欢字符串，给定字符串长度n，字符串中还必须包括k个不同的字符，相邻字符还不一样。输出符合这些条件的最小字符（按字典序排列）。很明显字符串前面肯定是ab的组合，在结尾递增的加上剩余字符即可。这个题要注意的地方有 除了1 1输出是a外，其他k == 1时输出都为-1，因为不能用一个字符组成一个前后字符不相等的字符串。（虽然是很浅显的特殊情况，但有时候就是容易忽略，今后要避免这样的情况发生）。
不过这个题我要抱怨一下，一开始编译的时候，str[i] = str[i - 1] + 1;这行代码得到的不是我想要的结果，除了前面的ab其他的打印出来都是笑脸，你难道想让我陪你玩一会在提交？笑什么笑。。。不过后来在演示这一奇葩现象的时候又恢复了正常了。亏我在代码中还用了强制转换，其实根本用不上好吗。
#include<stdio.h>
char str[];
int main()
{
int n,k,i,j,m;
scanf("%d%d",&n,&k);
if(n == && k==)printf("a\n");
else if(k > n || k==)printf("-1\n");
else if(k == n)
{
for(i = ;i < n;i++)
{
if(i == )
str[] = 'a';
else
str[i] = (char)((int)str[i-] +);
}
str[n] = '\0';
printf("%s\n",str);
}
else if(k < n)
{
m =n - (k - );
if(m % == )
{
for(i = ;i < m-;i = i+)
{
str[i] = 'a';
str[i+] = 'b';
}
for(;i < n;i++)
str[i] = (char)((int)str[i-] +);
}
else
{
for(i = ;i < m-;i = i+)
{
str[i] = 'a';
str[i+] = 'b';
}
str[i++] = 'a';
str[i++] = 'c';
for(;i < n;i++)
str[i] = (char)((int)str[i-] +);
}
str[n] = '\0';
printf("%s\n",str);
}
return ;
}
当然，这个题不用字符串存起来也行，可惜当时只想到用字符串的方法。
感觉自己写题解废话越来越多了，多写上了自己的感受，和wa的原因，主要还是因为写的这些东西都是为了留下纪念，所以废话多一点的话今后看起来比较亲切。若有路过的大神不要嫌弃啊（其实我这些渣题不会有大神来看的啦~）。
