A：题意：给一串字符串以abc和？组成，要求将？替换成abc，使得这个字符串相邻两个字母不等。
暴力模拟。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
string s;
int t;
int main()
{
t = read();
while (t--)
{
cin >> s;
int n = s.size();
bool flag = 1;
up(i, 1, n)
{
if (s[i] == s[i - 1]&&s[i]!='?')flag = 0;
}
if (!flag) { printf("%d\n", -1); }
else
{
if (n == 1)
{
if (s[0] == '?')s[0] = 'a';
}
else
{
up(i, 0, n)
{
if (i == 0)
{
if (s[i] == '?')
{
if (s[i + 1] == 'b')s[i] = 'a';
if (s[i + 1] == 'c')s[i] = 'b';
if (s[i + 1] == 'a')s[i] = 'c';
if (s[i + 1] == '?')s[i] = 'a';
}
}
else
{
if (s[i - 1] != '?'&&s[i] == '?')
{
if (i + 1 < n&&s[i + 1] == '?')
{
if (s[i - 1] == 'b')s[i] = 'a';
if (s[i - 1] == 'c')s[i] = 'b';
if (s[i - 1] == 'a')s[i] = 'c';
}
else if (i + 1 < n&&s[i + 1] != '?')
{
int a = 0, b = 0, c = 0;
if (s[i - 1] == 'a')a++;
if (s[i - 1] == 'b')b++;
if (s[i - 1] == 'c')c++;
if (s[i + 1] == 'a')a++;
if (s[i + 1] == 'b')b++;
if (s[i + 1] == 'c')c++;
if (a == 0)s[i] = 'a';
if (b == 0)s[i] = 'b';
if (c == 0)s[i] = 'c';
}
else if (i == n-1)
{
if (s[i - 1] == 'b')s[i] = 'a';
if (s[i - 1] == 'c')s[i] = 'b';
if (s[i - 1] == 'a')s[i] = 'c';
}
}
}
}
}
cout << s << endl;
}
}
return 0;
}B；题意：给一串permutation，遍历长度1-n，寻找每一个i，能否组成permutation。
思维，记录最左侧和最右侧，相减=len，ans++，因为这个时候1-i的所有数字必定全部在这个区间内。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 2e5 + 10;
int T, n;
int a[N];
int pos[N];
int main()
{
T = read();
while (T--)
{
n = read();
upd(i, 1, n)a[i] = read(),pos[a[i]]=i;
int lf = n + 1, rt = 0;
string ans = "";
upd(i, 1, n)
{
lf = min(lf, pos[i]);
rt = max(rt, pos[i]);
if (rt - lf + 1 == i)ans+='1';
else ans += '0';
}
cout << ans << endl;
}
}C：我又写了个复杂模拟。。。
题意：太复杂了。。
我的思路是，贪心。最后一行取拿不到牌子的，第一行取金牌，这样剩下的凑银牌和铜牌即可。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
int n;
int t;
const int N = 4e5 + 10;
int p[N];
int st[N];
int cnt[N];
int main()
{
t = read();
while (t--)
{
n = read();
int last = -1;
int top = 0;
upd(i, 1, n)
{
p[i] = read();
if (top == 0)
{
top++;
st[top] = p[i];
cnt[top]=1;
}
else {
if (st[top] != p[i])
{
st[++top] = p[i];
cnt[top] = 1;
}
else
{
cnt[top]++;
}
}
}
int half = n / 2;
if (top < 4)
{
printf("%d %d %d\n", 0, 0, 0);
}
else
{
int left = n - cnt[top];
int one = cnt[1];
int pos = top;
for (int i = top - 1; left > half&&i; i--)
{
left -= cnt[i];
pos = i;
}
int ans = left;
left -= one;
int twsum = 0;
int pos2 = pos;
for (int i = pos-1; twsum <= one&&i; i--)
{
twsum += cnt[i];
pos2 = i;
}
left -= twsum;
if (left > one)
{
printf("%d %d %d\n", one, left, twsum);
}
else
{
printf("%d %d %d\n", 0, 0, 0);
}
}
}
return 0;
}