mirror of
https://github.com/Vision-CAIR/MiniGPT-4.git
synced 2025-04-05 02:20:47 +00:00
150 lines
3.4 KiB
Python
150 lines
3.4 KiB
Python
# from math import factorial
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# def combine_number(n,k):
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# combinations = factorial(n) / factorial(n - k)
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# return combinations
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# # 定义一个函数来计算给定字符串中平衡串的子序列数量
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# def count_balanced_subsequences(s):
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# MOD = 10**9 + 7
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# # 初始化字母计数器
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# count = [0] * 26
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# for char in s:
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# count[ord(char) - ord('a')] += 1
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# result = 0
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# while sum(count)>0:
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# cnt = 0
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# for i in range(26):
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# if count[i] > 0:
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# cnt += 1
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# count[i] -= 1
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# print(count)
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# result += combine_number(cnt,2)
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# print(result)
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# return result
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# # 示例输入
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# n = 5
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# s = "ababc"
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# # 计算结果
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# result = count_balanced_subsequences(s)
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# print(result)
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# def calculate_scores(k, x, y):
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# # 根据题目描述，我们有以下等式：
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# # a + b + c = k
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# # c = a + x
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# # c = b - y
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# # 由上面两个等式可得：a + x = b - y
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# # 因此我们可以解这个线性方程组得到a, b, c的值
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# # a + (a + x) + (a + x + y) = k
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# # 3a + 2x + y = k
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# # 解得：
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# a = (k - 2*x - y) // 3
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# c = a + x
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# b = c + y
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# return a, b, c
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# # 示例输入
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# k, x, y = 441, 1, -20
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# # 计算输出
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# a, b, c = calculate_scores(k, x, y)
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# print(a, b, c)
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# def min_operations_to_equal(s, t):
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# operations = []
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# i = len(s) - 1
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# while i >= 0:
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# if s[i] != t[i]:
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# operations.append((1, i + 1, t[i]))
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# s = t[i] * (i + 1) + s[i + 1:] # 更新s字符串为操作后的状态
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# i -= 1
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# return operations
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# # 示例输入
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# s = "aabc"
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# t = "abcc"
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# # 计算所需的最小操作次数及具体操作
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# operations1 = min_operations_to_equal(s, t)
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# operations2 = min_operations_to_equal(t, s)
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# if len(operations1) < len(operations2):
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# operations = operations1
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# else:
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# operations = operations2
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# # 输出结果
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# print(len(operations))
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# for op in operations:
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# print(*op)
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# #include <iostream>
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# #include <linux/limits,h>
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# #include <string>
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# using namespace std;
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# struct node {
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# char a, b;
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# }ans[100100];
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# char col[9] = {'','a','b'
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# int main(){
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# string x; cin >>x;
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# char w=x[0],s=x[1];
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# int ww=w-'a'+ 1;
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# int ss =s -'0';
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# int tot = 0;
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# for(int i=1;i<=8; ++i){
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# if(1 != ss){
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# ans[++tot]=(node){w,(char)(i +'@')};
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# }
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# }
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# for(int i=1; i <= 8; ++i){
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# if(1 != ww){
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# ans[++tot]=(node){colli], s};
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# }
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# }
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# for(int i=1;i<8;++1){
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# for(int j=1;j<=8; ++j){
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# if(i+j==ww + ss){
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# if(i !=ss |lj != ww){
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# ans[++tot]=(node){col[j],(char)}
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# }
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# }
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# }
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# }
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# for(int i=1;i<=8;++i){
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# for(int j=1;j<=8; ++j){
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# if(i-j=ss-ww){
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# if(i !=ss ||j!= ww){
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# ans[++tot]=(node)fcol[j]，(char)
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# }
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# }
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# }
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# }
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# cout << tot.<< endl;
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# for(int i =1; i <= tot; ++i) cout << ans[i].a <<
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# return 0;
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# #include <linux/limits.h>
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# #include <string>using namespace stdj
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# const int N= 1001008
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# int n, a[N];
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# int main(){
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# cin >> n;
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# int tot1 = 0, tot0 = 0;
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# for(int i=1;i<= n; ++1){
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# cin >> a[i];
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# if(a[i]%2=0) tot0++;
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# else tot1++;
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# cout << abs(tot1-(n/2))<<endl;
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# return 0;
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