给出一个字符串，可以重新排列顺序，问所有的子串中abccca最多出现几次。
strs = input()
dic = {
'a': 0,
'b': 0,
'c': 0
}
for s in strs:
if s in dic:
dic[s] += 1
nums = list(dic.values())
print(min((nums[0] - 1), nums[1], int(nums[2]/3)))给出一个递增数组，要求找到一个点，使这个点到1号点和n号店的距离之差最小。
n = int(input())
nums = list(map(int, input().split()))
median = (nums[0]/2 + nums[n - 1]/2)
tmp = [0]*n
for i in range(n):
tmp[i] = abs(nums[i] - median)
r = min(tmp)
print(int(2*r))给出n个数，要求从中选任意个数，使得他们的和最大，且为7的倍数。
思路：dp[i][j], 记录前i个数的和对7取模为j的和的最大值即可。
n = int(input())
nums = list(map(int, input().split()))
MAX = int(1e6+7)
# create dp
dp = [[-float("inf")]*7 for _ in range(MAX)]
dp[0][0] = 0;
for i in range(1, n + 1):
for j in range(7):
# dp[i][j] is at least as large as dp[i - 1][j], that is, we can simply ignore the last number
dp[i][j] = max(dp[i][j], dp[i - 1][j])
dp[i][j] = max(dp[i][j], dp[i - 1][(j - nums[i - 1]) % 7] + nums[i - 1])
print((i,j,dp[i][j]))
print(dp[n][0])#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn= 1e6+7 ;
const int inf = 0x3f3f3f3f;
int n , a[maxn];
int dp[maxn][10];
int main(){
scanf("%d",&n);
for(int i = 1; i <= n; i ++){
scanf("%d",&a[i]);
}
memset(dp , -inf , sizeof(dp));
dp[0][0] = 0;
for(int i = 1; i <= n; i ++){
for(int j = 0; j < 7; j ++){
dp[i][j] = max(dp[i][j] , dp[i - 1][j]);
dp[i][j] = max(dp[i][j] , dp[i - 1][((j - a[i]) % 7 + 7) % 7] + a[i]);
//printf ("%d %d %d\n",i , j , dp[i][j]);
}
}
printf ("%d\n",dp[n][0]);
}给出长度为n的数组，求所有长度为奇数的子段的中位数之和。
思路：两个优先队列动态维护中位数 复杂度
from heapq import *
class MedianFinder:
def __init__(self):
"""
initialize your data structure here.
"""
self.A = [] # 小顶堆，保留较大的一半
self.B = [] # 大顶堆，保留较小的一半
def addNum(self, num: int) -> None:
if len(self.A) != len(self.B):
heappush(self.A, num)
heappush(self.B, -heappop(self.A))
else:
heappush(self.B, -num)
heappush(self.A, -heappop(self.B))
def findMedian(self) -> float:
return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0
n = int(input())
nums = list(map(int, input().split()))
ans = 0
for l in range(n):
m = MedianFinder()
m.addNum(nums[l])
ans += m.findMedian()
for r in range(l + 1, n):
m.addNum(nums[r])
if (r - l + 1)%2:
ans += m.findMedian()
print(ans)#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn= 1e6+7 ;
const int inf = 0x3f3f3f3f;
int n , a[maxn];
priority_queue<int>q1,q2;
void init(){
while(q1.size()) q1.pop();
while(q2.size()) q2.pop();
}
void add(int now){
if(q2.size() == 0){
q2.push(-now);
return ;
}
if(now >= -q2.top()){
q2.push(-now);
if(q2.size() > q1.size() + 1){
int x = q2.top();
q2.pop();
q1.push(-x);
}
}
else{
q1.push(now);
if(q1.size() > q2.size()){
int x = q1.top();
q1.pop();
q2.push(-x);
}
}
}
int main(){
scanf("%d",&n);
ll ans = 0;
for(int i = 1; i <= n; i ++){
scanf("%d",&a[i]);
ans += a[i];
}
for(int l = 1; l <= n; l ++){
init();
add(a[l]);
for(int r = l + 1;r <= n; r ++){
add(a[r]);
if((r - l + 1) % 2){
ans -= q2.top();
}
}
}
printf ("%lld\n",ans);
}有n个任务，完成每个任务需要的时间为l[i]l[i],在完成该任务前，必须先完成一些前置任务，多个任务可以同时进行，问每个任务的最早完成时间。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn= 1e6+7 ;
const int inf = 0x3f3f3f3f;
int n;
int l[maxn] , now[maxn] , de[maxn],ans , vis[maxn];
vector<int> vec[maxn] , e[maxn];
void topo(){
queue<int> que;
for(int i = 1; i <= n; i ++){
if(de[i] == 0){
que.push(i);
vis[i] = 1;
now[i] = now[i] + l[i];
}
}
while(que.size()){
int u = que.front();
que.pop();
for(int i = 0; i < e[u].size(); i ++){
int to = e[u][i];
de[to]--;
now[to] = max(now[to] , now[u]);
if(de[to] == 0 && vis[to] == 0){
que.push(to);
vis[to] = 1;
now[to] += l[to];
}
}
}
}
int main(){
scanf("%d",&n);
for(int i = 1; i <= n; i ++){
scanf("%d",&l[i]);
int c , x;
scanf("%d",&c);
de[i] = c;
for(int j = 1; j <= c; j ++){
scanf("%d",&x);
vec[i].push_back(x);
e[x].push_back(i);
}
}
topo();
for(int i = 1; i <= n; i ++){
printf ("%d ",now[i]);
}
}梯度下降与最速下降