比赛链接：http://codeforces.com/contest/550
A. Two Substrings
You are given string s. Your task is to determine if the given strings contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order).
The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters.
Print "YES" (without the quotes), if strings contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise.
ABA
NO
BACFAB
YES
AXBYBXA
NO
In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings:BACFAB.
In the third sample test there is no substring "AB" nor substring "BA".
题目大意：给一个字符串，问是否能找到两个不相互覆盖的子串"AB"和"BA"
题目分析：暴力扫4次，第一二次先扫"AB"后扫"BA"。第三四次先扫"BA"后扫"AB"，注意这组例子ABACCCAB
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 1e5 + 5;
char s1[MAX], s2[MAX];
int main()
{
scanf("%s", s1);
memcpy(s2, s1, sizeof(s1));
int len = strlen(s1);
bool f1 = false;
bool f2 = false;
for(int i = 0; i < len; i++)
{
if(s1[i] == 'A' && s1[i + 1] == 'B')
{
f1 = true;
s1[i] = s1[i + 1] = '*';
break;
}
}
for(int i = 0; i < len; i++)
if(s1[i] == 'B' && s1[i + 1] == 'A')
f2 = true;
if(f1 && f2)
{
printf("YES
");
return 0;
}
f1 = f2 = false;
for(int i = 0; i < len; i++)
{
if(s2[i] == 'B' && s2[i + 1] == 'A')
{
f1 = true;
s2[i] = s2[i + 1] = '*';
break;
}
}
for(int i = 0; i < len; i++)
if(s2[i] == 'A' && s2[i + 1] == 'B')
f2 = true;
if(f1 && f2)
{
printf("YES
");
return 0;
}
printf("NO
");
}
You have n problems. You have estimated the difficulty of thei-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at leastl and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at leastx.
Find the number of ways to choose a problemset for the contest.
The first line contains four integers n,l, r,x (1 ≤ n ≤ 15,1 ≤ l ≤ r ≤ 109,1 ≤ x ≤ 106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 106) — the difficulty of each problem.
Print the number of ways to choose a suitable problemset for the contest.
3 5 6 1 1 2 3
2
4 40 50 10 10 20 30 25
2
5 25 35 10 10 10 20 10 20
6
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
题目大意：有n个问题每一个难度为ci。要选一些题出来，要求这些题的总难度不超过r且不小于l。且难度最大的和难度最小的差不能小于x，问有多少种选法
题目分析：n等于15。随便搞，我是直接深搜，简洁明了
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const INF = 0x3fffffff;
int n, l, r, x;
int c[20];
int ans;
void DFS(int idx, int ma, int mi, int sum) //第几个题。当前最大。当前最小。难度总和
{
if(idx == n + 1)
return;
if(sum <= r && sum >= l && x <= ma - mi && idx == n)
ans ++;
DFS(idx + 1, max(ma, c[idx]), min(mi, c[idx]), sum + c[idx]); //选第idx个
DFS(idx + 1, ma, mi, sum); //不选第idx个
return;
}
int main()
{
scanf("%d %d %d %d", &n, &l, &r, &x);
for(int i = 0; i < n; i++)
scanf("%d", &c[i]);
ans = 0;
DFS(0, 0, INF, 0);
printf("%d
", ans);
}C. Divisibility by Eight
You are given a non-negative integer n, its decimal representation consists of at most 100 digits and doesn't contain leading zeroes.
Your task is to determine if it is possible in this case to remove some of the digits (possibly not remove any digit at all) so that the result contains at least one digit, forms a non-negative integer, doesn't have leading zeroes and is divisible by 8. After the removing, it is forbidden to rearrange the digits.
If a solution exists, you should print it.
The single line of the input contains a non-negative integern. The representation of number n doesn't contain any leading zeroes and its length doesn't exceed100 digits.
Print "NO" (without quotes), if there is no such way to remove some digits from numbern.
Otherwise, print "YES" in the first line and the resulting number after removing digits from numbern in the second line. The printed number must be divisible by8.
If there are multiple possible answers, you may print any of them.
3454
YES
344
10
YES
0
111111
NO
题目大意：给个数字，问删去当中一些数字后能不能被8整除
题目分析：打了个小表，找了找规律，首先有8或0的直接能够，其次包括16。24，32。56。64，72。96之中的一个的都能够。再其次，先找一个奇数，它后面为12，36。44，52。76，92之中的一个的都能够，数字长度非常小，暴搞
#include <cstdio>
#include <cstring>
int const MAX = 105;
char s[MAX];
int len;
// 0
// 8
// 16
// 24
// 32
// 56
// 64
// 72
// 96
// 112
// 136
// 144
// 152
// 176
// 192
// 312
// 336
bool judge1(int i, char ch1, char ch2)
{
if(s[i] == ch1)
{
for(int j = i + 1; j < len; j++)
{
if(s[j] == ch2)
{
printf("YES
%c%c
", ch1, ch2);
return true;
}
}
}
return false;
}
bool judge2(char ch, int i, char ch1, char ch2)
{
for(int j = i + 1; j < len; j++)
{
if(s[j] == ch1)
{
for(int k = j + 1; k < len; k ++)
{
if(s[k] == ch2)
{
printf("YES
%c%c%c
", ch, ch1, ch2);
return true;
}
}
}
}
return false;
}
int main()
{
scanf("%s", s);
len = strlen(s);
for(int i = 0; i < len; i++)
{
if(s[i] == '0')
{
printf("YES
0
");
return 0;
}
if(s[i] == '8')
{
printf("YES
8
");
return 0;
}
if(judge1(i, '1', '6'))
return 0;
if(judge1(i, '2', '4'))
return 0;
if(judge1(i, '3', '2'))
return 0;
if(judge1(i, '5', '6'))
return 0;
if(judge1(i, '6', '4'))
return 0;
if(judge1(i, '7', '2'))
return 0;
if(judge1(i, '9', '6'))
return 0;
if((s[i] - '0') % 2 == 1 && judge2(s[i], i, '1', '2'))
return 0;
if((s[i] - '0') % 2 == 1 && judge2(s[i], i, '3', '6'))
return 0;
if((s[i] - '0') % 2 == 1 && judge2(s[i], i, '4', '4'))
return 0;
if((s[i] - '0') % 2 == 1 && judge2(s[i], i, '5', '2'))
return 0;
if((s[i] - '0') % 2 == 1 && judge2(s[i], i, '7', '6'))
return 0;
if((s[i] - '0') % 2 == 1 && judge2(s[i], i, '9', '2'))
return 0;
}
printf("NO
");
}
Implication is a function of two logical arguments, its value is false if and only if the value of the first argument is true and the value of the second argument is false.
Implication is written by using character '
',
and the arguments and the result of the implication are written as '0' (false) and '1' (true). According
to the definition of the implication:
When a logical expression contains multiple implications, then when there are no brackets, it will be calculated from left to fight. For example,
.
When there are brackets, we first calculate the expression in brackets. For example,
.
For the given logical expression 
determine if it is possible to place there brackets
so that the value of a logical expression is false. If it is possible, your task is to find such an arrangement of brackets.
The first line contains integer n (1 ≤ n ≤ 100 000) — the number of arguments in a logical expression.
The second line contains n numbersa1, a2, ..., an
(
), which means the values of arguments in the expression in the order they occur.
Print "NO" (without the quotes), if it is impossible to place brackets in the expression so that its value was equal to 0.
Otherwise, print "YES" in the first line and the logical expression with the required arrangement of brackets in the second line.
The expression should only contain characters '0', '1', '-' (character with ASCII code 45), '>' (character with ASCII code 62), '(' and ')'. Characters '-' and '>' can occur in an expression only paired like that: ("->") and represent implication. The total number of logical arguments (i.e. digits '0' and '1') in the expression must be equal to n. The order in which the digits follow in the expression from left to right must coincide witha1, a2, ..., an.
The expression should be correct. More formally, a correct expression is determined as follows:
- Expressions "0", "1" (without the quotes) are correct.
- If v1,v2 are correct, thenv1->v2 is a correct expression.
- If v is a correct expression, then(v) is a correct expression.
The total number of characters in the resulting expression mustn't exceed106.
If there are multiple possible answers, you are allowed to print any of them.
4
0 1 1 0
YES
(((0)->1)->(1->0))
2
1 1
NO
1
0
YES
0
题目分析：从给的四个式子中能够发现假设结果要为0，最后一位必须是0。如今要做的就是再最后一个0之前构造1，我们能够发现假设最后一个0的前面一个是1，那么无论这个1之前是什么最后答案都是1。由于0 ->1=1，1 ->1=1，即与前面的值无关，所以我们转过来考虑不可能的情况，考虑最后一个0的前面是0，由于要构1。又0->0=1，1->0=0也就是说这个0前面不能是1，依次类推能够得到。假设倒数第二个0的前面都是1，那必定无解。否则就有解
#include <cstdio>
#include <cstring>
int const MAX = 1e6 + 5;
int n, a[MAX];
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
if(n == 1) //特判一个数
{
if(a[1] == 1)
printf("NO
");
else
printf("YES
0
");
return 0;
}
if(a[n] == 1)
{
printf("NO
");
return 0;
}
bool flag = false;
for(int i = 1; i <= n - 2; i++)
{
if(a[i] != 1)
{
flag = true;
break;
}
}
if(!flag && a[n - 1] == 0 && a[n] == 0) //1111111 1 0Y 0 0N
{
printf("NO
");
return 0;
}
printf("YES
");
for(int i = 1; i <= n - 2; i++)
printf("(%d->", a[i]);
printf("%d", a[n - 1]);
for(int i = 1; i <= n - 2; i++)
printf(")");
printf("->0
");
}
D也是个构造，没来及补