A. Dreamoon and Stairs
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
dreamoon wants to climb up a stair of n steps. he can climb 1 or 2 steps at each move. dreamoon wants the number of moves to be a multiple of an integer m.
What is the minimal number of steps making him climb to the top of the stairs that satisfies his condition?
Input
The single line contains two space separated integers n, m (0?
Output
Print a single integer ? the minimal number of moves being a multiple of m. If there is no way he can climb satisfying condition print ?-?1 instead.
Sample test(s)
Input
10 2
Output
Input
3 5
Output
-1
Note
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
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For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5.
题意：给两个数n,m，
现在要求爬楼梯，一次可以爬
一个或两个阶梯，要求爬完n个阶梯，
求爬的次数为m的倍数时，最少次数是多少，
若不存在答案输出-1。
做法：很明显尽量爬两步。
代码：
#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;int main(){ int n,m,x,y; cin>>n>>m; x=n/2+n%2; y=n/2; while(y>-1) { if(x%m==0) { cout<<x<<endl; return 0; } x++; y--; } cout<<-1<<endl;}
B. Dreamoon and WiFi
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Dreamoon is standing at the position 0 on a number line. Drazil is sending a list of commands through Wi-Fi to Dreamoon's smartphone and Dreamoon follows them.
Each command is one of the following two types:
- Go 1 unit towards the positive direction, denoted as '+'
- Go 1 unit towards the negative direction, denoted as '-'
But the Wi-Fi condition is so poor that Dreamoon's smartphone reports some of the commands can't be recognized and Dreamoon knows that some of them might even be wrong though successfully recognized. Dreamoon decides to follow every recognized command and toss a fair coin to decide those unrecognized ones (that means, he moves to the 1 unit to the negative or positive direction with the same probability 0.5).
You are given an original list of commands sent by Drazil and list received by Dreamoon. What is the probability that Dreamoon ends in the position originally supposed to be final by Drazil's commands?
Input
The first line contains a string s1 ? the commands Drazil sends to Dreamoon, this string consists of only the characters in the set {'+', '-'}.
The second line contains a string s2 ? the commands Dreamoon's smartphone recognizes, this string consists of only the characters in the set {'+', '-', '?'}. '?' denotes an unrecognized command.
Lengths of two strings are equal and do not exceed 10.
Output
Output a single real number corresponding to the probability. The answer will be considered correct if its relative or absolute error doesn't exceed 10?-?9.
Sample test(s)
Input
++-+-+-+-+
Output
1.000000000000
Input
+-+-+-??
Output
0.500000000000
Input
+++??-
Output
0.000000000000
Note
For the first sample, both s1 and s2 will lead Dreamoon to finish at the same position ?+?1.
For the second sample, s1 will lead Dreamoon to finish at position 0, while there are four possibilites for s2: {"+-++", "+-+-", "+--+", "+---"} with ending position {+2, 0, 0, -2} respectively. So there are 2 correct cases out of 4, so the probability of finishing at the correct position is 0.5.
For the third sample, s2 could only lead us to finish at positions {+1, -1, -3}, so the probability to finish at the correct position ?+?3 is 0.
题意：给两个长度不超过10的相同长度的串，第一个串由+和-组成，
第二个串由+和-和？组成，
初始状态为0，+代表+1，-代表-1，
现在要求上下结果状态相同时，概率是多少。
做法：先判断是否有解，若无解输出0，
若有解，求出第二个串还需要多少个+，
统计？的数量，求出组合数除以所有的可能数量。
或者，直接爆搜，因为最多只有10个字符，
所以，不会超时。
代码：
#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;int s1,s2,s3;void dfs(int step,int val){ if(step==s2) { if(val==s1) s3++; return; } dfs(step+1,val+1); dfs(step+1,val-1);}int main(){ string a,b; int n,i; cin>>a>>b; n=a.length(); s1=0; for(i=0;i<n;i++) if(a[i]=='+') s1++; else s1--; s2=0; for(i=0;i<n;i++) if(b[i]=='+') s1--; else if(b[i]=='-') s1++; else s2++; s3=0; dfs(0,0); printf("%.10f",(double)s3/(1<<s2));}
C. Dreamoon and Sums
time limit per test
1.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occasionally. He wants to calculate the sum of all nice integers. Positive integer x is called nice if and , where k is some integer number in range [1,?a].
By we denote the quotient of integer division of x and y. By we denote the remainder of integer division of x and y. You can read more about these operations here: http://goo.gl/AcsXhT.
The answer may be large, so please print its remainder modulo 1?000?000?007 (109?+?7). Can you compute it faster than Dreamoon?
Input
The single line of the input contains two integers a, b (1?≤?a,?b?≤?107).
Output
Print a single integer representing the answer modulo 1?000?000?007 (109?+?7).
Sample test(s)
Input
1 1
Output
Input
2 2
Output
Note
For the first sample, there are no nice integers because is always zero.
For the second sample, the set of nice integers is {3,?5}.
题意：给出两个数a,b,当div(x,b)/mod(x,b)=k，1
所有可能的x的和对1?000?000?007取余的结果。
做法：
设y=div(x,b),z=mod(x,b)，
可以得到
y=z*k,y*b+z=x,联立得
(kb+1)z=x,
下面用到求和公式，
然后假设k为常量，得到x=b(b-1)*(kb+1)/2，
最后k还原为变量，得到x=b(b-1)/2*[(1+a)a*b/2+a]
代码：
#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;const long long mod=1000000007;int main(){ long long a,b,t1,t2,t3; cin>>a>>b; t1=(1+a)*a/2%mod; t1=(t1*b)%mod; t1=(t1+a)%mod; t2=b*(b-1)/2%mod; t3=(t1*t2)%mod; cout<<t3;}
D. Dreamoon and Sets
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Dreamoon likes to play with sets, integers and . is defined as the largest positive integer that divides both a and b.
Let S be a set of exactly four distinct integers greater than 0. Define S to be of rank k if and only if for all pairs of distinct elements si, sj from S, .
Given k and n, Dreamoon wants to make up n sets of rank k using integers from 1 to m such that no integer is used in two different sets (of course you can leave some integers without use). Calculate the minimum m that makes it possible and print one possible solution.
Input
The single line of the input contains two space separated integers n, k (1?≤?n?≤?10?000,?1?≤?k?≤?100).
Output
On the first line print a single integer ? the minimal possible m.
On each of the next n lines print four space separated integers representing the i-th set.
Neither the order of the sets nor the order of integers within a set is important. If there are multiple possible solutions with minimal m, print any one of them.
Sample test(s)
Input
1 1
Output
51 2 3 5
Input
2 2
Output
222 4 6 2214 18 10 16
Note
For the first example it's easy to see that set {1,?2,?3,?4} isn't a valid set of rank 1 since .
题意：求出n个集合均为4个元素，且每个集合内任意两两
元素的最大公约数为k，集合不允许有交集，
当n*4个元素最大值最小时，输出所有n个集合，
如果有多组结果输出任意一组。
做法：
很明显，当任意集合内的所有元素除以k后，
两两元素之间是互质的，然后为了满足元素最大值最小，
在除掉k后，任意集合内肯定是由3个奇数，1个偶数组成，
因为若少一个奇数，就会有至少一对数不互质，
若多一个奇数，元素最大值就会变大。
所以，从1开始构建所有元素集合即可。
代码：
#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;int main(){ int n,k; cin>>n>>k; cout<<(6*n-1)*k<<endl; while(n--) { cout<<(6*n+1)*k<<" "<<(6*n+2)*k<<" "<<(6*n+3)*k<<" "<<(6*n+5)*k<<endl; }}
E. Dreamoon and Strings
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. He wants to make this number as big as possible.
More formally, let's define as maximum value of over all s' that can be obtained by removing exactly x characters from s. Dreamoon wants to know for all x from 0 to |s| where |s| denotes the length of string s.
Input
The first line of the input contains the string s (1?≤?|s|?≤?2?000).
The second line of the input contains the string p (1?≤?|p|?≤?500).
Both strings will only consist of lower case English letters.
Output
Print |s|?+?1 space-separated integers in a single line representing the for all x from 0 to |s|.
Sample test(s)
Input
aaaaaaa
Output
2 2 1 1 0 0
Input
axbaxxbab
Output
0 1 1 2 1 1 0 0
Note
For the first sample, the corresponding optimal values of s' after removal 0 through |s|?=?5 characters from s are {"aaaaa", "aaaa", "aaa", "aa", "a", ""}.
For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab", "a", ""}.
题意：给出两个串，当对第一个串长度为n时，
分别除去1到n个字符可以获得n个新串，
要求每个新串的子串为第二个串的数量最大，
输出n个新串的所谓最大数量。
做法：
dp[i][j]:第一个串0到i的位置除去j个字符可以得到的最大数量
（j a[i]：从i开始，包含第二个串的最近位置，若没有赋值为-1，
ns:第一个串的长度
np:第二个串的长度
有
dp[i+1][j]=max(dp[i][j],dp[i+1][j]);
//若不删除第i个字符dp[i+1][j]=dp[i][j]
dp[i+1][j+1]=max(dp[i][j],dp[i+1][j+1]);
//若删除第i个字符dp[i+1][j+1]=dp[i][j]
if(a[i]>0)
dp[i+a[i]][j+a[i]-np]=max(dp[i][j]+1,dp[i+a[i]][j+a[i]-np]);
//若删除从i开始后到i+a[i]不包含与第二个串相同的字符，
dp[i+a[i]][j+a[i]-np]=dp[i][j]+1
因为dp[ns-1][j]木有考虑ns-1个字符是否删除的情况，
所以最终答案存在dp[ns][j]中。
代码：
#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;int a[2010],dp[2010][2010];int main(){ string s,p; int i,j,k,ns,np; cin>>s>>p; ns=s.length(); np=p.length(); for(i=0;i<ns;i++) { j=0;k=i; while(j<np&&k<ns) { if(p[j]==s[k]) j++; k++; } a[i]= j==np?k-i:-1; } for(i=0;i<ns;i++) for(j=0;j<=i;j++) { dp[i+1][j]=max(dp[i][j],dp[i+1][j]); dp[i+1][j+1]=max(dp[i][j],dp[i+1][j+1]); if(a[i]>0) dp[i+a[i]][j+a[i]-np]=max(dp[i][j]+1,dp[i+a[i]][j+a[i]-np]); } i=ns; cout<<dp[i][0]; for(j=1;j<=ns;j++) cout<<" "<<dp[i][j];} 
