#个人赛第四场解题总结#
涉及字符串的题比较多，有一道题原先在codeforces做过，
【题解】
Description
A and B are preparing themselves for programming contests.
To train their logical thinking and solve problems better, A and B decided to play chess. During the game A wondered whose position is now stronger.
For each chess piece we know its weight:
- the queen's weight is 9,
- the rook's weight is 5,
- the bishop's weight is 3,
- the knight's weight is 3,
- the pawn's weight is 1,
- the king's weight isn't considered in evaluating position.
The player's weight equals to the sum of weights of all his pieces on the board.
As A doesn't like counting, he asked you to help him determine which player has the larger position weight.
Input
The input contains eight lines, eight characters each — the board's description.
The white pieces on the board are marked with uppercase letters, the black pieces are marked with lowercase letters.
The white pieces are denoted as follows: the queen is represented is 'Q', the rook — as 'R', the bishop — as'B', the knight — as 'N', the pawn — as 'P', the king — as 'K'.
The black pieces are denoted as 'q', 'r', 'b', 'n', 'p', 'k', respectively.
An empty square of the board is marked as '.' (a dot).
It is not guaranteed that the given chess position can be achieved in a real game. Specifically, there can be an arbitrary (possibly zero) number pieces of each type, the king may be under attack and so on.
Output
Print "White" (without quotes) if the weight of the position of the white pieces is more than the weight of the position of the black pieces, print "Black" if the weight of the black pieces is more than the weight of the white pieces and print "Draw" if the weights of the white and black pieces are equal.
Sample Input
........
........
........
........
........
........
...rk...
pppppppp
........
........
........
........
PPPPPPPP
RNBQKBNR
...k....
........
........
........
........
K...Q...
........
Hint
In the first test sample the weight of the position of the white pieces equals to 9, the weight of the position of the black pieces equals 5.
In the second test sample the weights of the positions of the black and the white pieces are equal to 39.
In the third test sample the weight of the position of the white pieces equals to 9, the weight of the position of the black pieces equals to 16.
题目所述略长，其实看懂了非常容易理解：
统计大写字母和小写字母的对应的数字值，最后判断关系即可：
代码：
#include <iostream>
#include <stdio.h>
#include <map>
#include <stack>
#include <math.h>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn=10005;
#define mem(a,b) memset(a,b,sizeof(a))
#define Max(a,b) a>b?a:b;
#define Min(a,b) a<b?a:b;
char mapp[maxn][1005];
int main()
{
int n=0,m=0,i,j;
for(i=0;i<8;i++)
{
for(j=0;j<8;j++)
{
scanf("%c",&mapp[i][j]);
}
getchar();
}
for(i=0;i<8;i++)
for(j=0;j<8;j++)
{
if(mapp[i][j]=='Q') m+=9;
if(mapp[i][j]=='R') m+=5;
if(mapp[i][j]=='B') m+=3;
if(mapp[i][j]=='N') m+=3;
if(mapp[i][j]=='P') m+=1;
if(mapp[i][j]=='q') n+=9;
if(mapp[i][j]=='r') n+=5;
if(mapp[i][j]=='b') n+=3;
if(mapp[i][j]=='n') n+=3;
if(mapp[i][j]=='p') n+=1;
}
if(m==n)
{
puts("Draw");
}
if(m>n) {puts("White");}
if(m<n) {puts("Black");}
}
Description
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 ≤ n ≤ 105) — the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers с1, с2, ..., сn - 2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Sample Input
1 5 8 123 7
123 7 5 1
5 1 7
123
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
3
Hint
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
题目看数据就懂了，只是不知道为什么这么出，首先想到的是求和相减，（或者直接模拟一遍）但是应该不会这么简单，感觉要用容器或者异或，但是实际可能是想难了。
代码：
#include <iostream>
#include <stdio.h>
#include <map>
#include <stack>
#include <math.h>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn=10005;
#define mem(a,b) memset(a,b,sizeof(a))
#define Max(a,b) a>b?a:b;
#define Min(a,b) a<b?a:b;
char mapp[maxn][1005];
int a[1000000],b[1000000],c[1000000];
int main()
{
mem(a,0),mem(b,0),mem(c,0);
int n,m,i,j,k,l;
scanf("%d",&n);
int a1=0,a2=0,a3=0;
for(i=0; i<n; i++)
{
scanf("%d",&a[i]);
a1+=a[i];
}
for(i=0; i<n-1; i++)
{
scanf("%d",&b[i]);
a2+=b[i];
}
for(i=0; i<n-2; i++)
{
scanf("%d",&c[i]);
a3+=c[i];
}
printf("%d\n",a1-a2);
printf("%d\n",a2-a3);
}
Description
A and B are preparing themselves for programming contests.
An important part of preparing for a competition is sharing programming knowledge from the experienced members to those who are just beginning to deal with the contests. Therefore, during the next team training A decided to make teams so that newbies are solving problems together with experienced participants.
A believes that the optimal team of three people should consist of one experienced participant and two newbies. Thus, each experienced participant can share the experience with a large number of people.
However, B believes that the optimal team should have two experienced members plus one newbie. Thus, each newbie can gain more knowledge and experience.
As a result, A and B have decided that all the teams during the training session should belong to one of the two types described above. Furthermore, they agree that the total number of teams should be as much as possible.
There are n experienced members and m newbies on the training session. Can you calculate what maximum number of teams can be formed?
Input
The first line contains two integers n and m (0 ≤ n, m ≤ 5·105) — the number of experienced participants and newbies that are present at the training session.
Output
Print the maximum number of teams that can be formed.
Sample Input
Hint
Let's represent the experienced players as XP and newbies as NB.
In the first test the teams look as follows: (XP, NB, NB), (XP, NB, NB).
In the second test sample the teams look as follows: (XP, NB, NB), (XP, NB, NB), (XP, XP, NB).
题目看懂就明白了：
给出 n 个 experienced participants 和 m 个 newbies ，需要组成尽量多的组，组由3个人组成。有两种组合方式：（1）1 个 experienced participant 和 2 个 newbie （2）2 个 experienced participant 和 1 个 newbie。问最多能组成的组数是多少组。 （排列组合）
如果大的数不比小的数达到两倍以上的话，即满足 ，
ps:这样写比较直观
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
int n, m;
scanf("%d %d",&n,&m);
if(m>=2*n)
{
cout<<n<<endl;
}
else if(n>=2*m)
{
cout<<m<<endl;
}
else{
cout<<(m+n)/3<<endl;
}
return 0;
}
Description
Vasya became interested in bioinformatics. He's going to write an article about similar cyclic DNA sequences, so he invented a new method for determining the similarity of cyclic sequences.
Let's assume that strings s and t have the same length n, then the function h(s, t) is defined as the number of positions in which the respective symbols of s and t are the same. Function h(s, t) can be used to define the function of Vasya distance ρ(s, t):
where
is obtained from string
s, by applying left circular shift
i times. For example,
ρ("AGC", "CGT") =
h("AGC", "CGT") + h("AGC", "GTC") + h("AGC", "TCG") +
h("GCA", "CGT") + h("GCA", "GTC") + h("GCA", "TCG") +
h("CAG", "CGT") + h("CAG", "GTC") + h("CAG", "TCG") =
1 + 1 + 0 + 0 + 1 + 1 + 1 + 0 + 1 = 6
Vasya found a string s of length n on the Internet. Now he wants to count how many strings t there are such that the Vasya distance from the string s attains maximum possible value. Formally speaking, t must satisfy the equation: 
.
Vasya could not try all possible strings to find an answer, so he needs your help. As the answer may be very large, count the number of such strings modulo 109 + 7.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 105).
The second line of the input contains a single string of length n, consisting of characters "ACGT".
Output
Print a single number — the answer modulo 109 + 7.
Sample Input
C
AG
TTT
Hint
Please note that if for two distinct strings t1 and t2 values ρ(s, t1) и ρ(s, t2) are maximum among all possible t, then both strings must be taken into account in the answer even if one of them can be obtained by a circular shift of another one.
In the first sample, there is ρ("C", "C") = 1, for the remaining strings t of length 1 the value of ρ(s, t) is 0.
In the second sample, ρ("AG", "AG") = ρ("AG", "GA") = ρ("AG", "AA") = ρ("AG", "GG") = 4.
In the third sample, ρ("TTT", "TTT") = 27
之前有印象，
题意：给一个由A，T，C，G组成的字符串s，求使函数ρ(s,t)最大的t的个数。易知，t应该由字符串s中个数最多的字母组成（总数最多的字母只有一个的时候很好理解，如果有多个字母的个数相同，那么就可以将这些个数相同的字母看成一个字母，这样是等价的），求出字母个数最多的种类数x，根据排列组合，答案就是x^n。
代码：
#include <iostream>
#include <stdio.h>
#include <map>
#include <stack>
#include <math.h>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn=100100;
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define Max(a,b) a>b?a:b;
#define Min(a,b) a<b?a:b;
char mapp[maxn][1005];
int a[10],b[100000];
const int mod=1e9+7;
int n,m,i,j,A,G,C,T;
char str[maxn];
int main()
{
// while(scanf("%d",&n))
// {
scanf("%d",&n);
getchar();
scanf("%s",str);
int ss=0;
for(int i=0; i<n; i++)
{
if(str[i]=='A') A++;
if(str[i]=='G') G++;
if(str[i]=='C') C++;
if(str[i]=='T') T++;
}
int m1=Max(A,G);
int m2=Max(C,T);
int maxx=Max(m1,m2);
if(A==maxx) ss++;
if(G==maxx) ss++;
if(C==maxx) ss++;
if(T==maxx) ss++;
LL ans=1;
for(int i=0; i<n; i++)
{
ans=(ans*ss)%mod;
}
printf("%lld\n",ans%mod);
}
E - 久违的月赛之一和
I - 久违的月赛之二，实在没思路。不知道题目要我们求的是什么，赛后看了解题报告：似乎一知半解：
参考博客： http://cache.baiducontent.com/c?m=9f65cb4a8c8507ed4fece7631046893b4c4380146d96864968d4e414c42246111c3aa6e07b22121980853a3c50f11e41bca770216c5d61aa98cf8a4ad6bc912d3bcd7a742613d20955c418dfdc4653d620e15aeaae12&p=8b2a9700959459b908e2962856&newp=9a7e8614ce904ead17bd9b7c4453d8304a02c70e3f98&user=baidu&fm=sc&query=fzu-+2139&qid=a3324cd5000853b1&p1=1
大家如果有问题，想法，欢迎交流~~