学会使用sort函数进行灵活判断
1.确定字符串是否包含唯一字符
cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
string s;
cin>>s;
sort(s.begin(),s.end());
int flag = 1;
int n = s.size();
for(int i = 0; i < n; i++){
if(s[i]==s[i+1]) flag = 0;
}
if(flag){
cout<<"YES";
}
else{
cout<<"NO";
}
return 0;
}一、进制的转换
eg1
cpp
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 50;
int a[N];
int main(){
string s = "2021ABCD";
int n = s.length();
for(int i = 0; i < n; i++){
if('0' <= s[i] && s[i] <= '9') a[i] = s[i] - '0';
else a[i] = s[i] - 'A' + 10;
}
ll res = 0;
for(int i = 0 ;i < n; i++){
res = res * 16 + a[i];
}
cout << res << '\n';
return 0;
}或者使用0x2021ABCD
cpp
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
int main(){
ll res = 0x2021ABCD;
cout<<res;
return 0;
}归纳可得到任意进制转换为十进制的通用解法
eg2
cpp
#include<bits/stdc++.h>
using namespace std;
const int N = 50;
int a[N];
using ll = long long;
int main(){
string s = "2022";
int n = s.length();
for(int i = 0; i < n; i++){
a[i] = s[i] - '0';
}
ll res = 0;
for(int i = 0; i < n; i++){
res = res * 9 + a[i];
}
cout<<res<<'\n';
return 0;
}eg3 N进制转换为M进制
cpp
#include<bits/stdc++.h>
using namespace std;
using ll = long long ;
const int N = 1000;
int a[N];
char ch[] = {'0', '1','2', '3','4', '5', '6', '7', '8', '9', 'A', 'B','C','D','E','F'};
void fun(){
int n,m;
string s;
cin>> n>> m;
cin>>s;
int len = s.length();
for(int i = 0; i < len; i ++){
if('0' <= s[i]&& s[i] >= '9') a[i] = s[i] - '0';
else a[i] = s[i] - 'A' + 10;
}
ll res = 0;
for(int i = 0; i < n; i++){
res = res * n + a[i];
}
string st;
while(res){
st += ch[res % m];
res /= m;
}
reverse(st.begin(),st.end());
cout<< st <<'\n';
return ;
}
int main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int n;
cin>>n;
for(int i = 0; i < n; i++){
fun();
}
return 0;
}我们可以得到十进制转为其他进制的解法以及任意进制转换为任意进制的通用结论
二、前缀和
cpp
#include<bits/stdc++.h>
using namespace std;
using ll = long long ;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
ll a[N];
ll pf[N][6];//前缀和数组
int n, m, l ,r, k;
int main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
cin>> n >> m;
for(int i = 1 ;i <= n; i++){
cin >> a[i];
for(int j = 1; j <= 5 ; j++){
pf[i][j] = pf[i - 1][j] + pow(a[i],j);
}
}
while(m--){
cin>>l>>r>>k;
cout << (pf[r][k]-pf[l-1][k] + mod) % mod << '\n';//先+mod是为了处理负数的情况
}
return 0;
}
cpp
#include<bits/stdc++.h>
using namespace std;
const int N = 1005;
int main(){
string s;
cin>> s;
int n = s.length();
vector<int> pf(n+1,0);
for(int i = 1 ; i <= n; i++){
if(s[i - 1] == 'L') pf[i] = pf[i - 1] + 1;
else pf[i] = pf[i - 1] - 1;
}
int ans = 0;
for(int i = 1; i <= n; i++){
for(int j = i; j <= n; j++){
if(pf[j] - pf[i - 1] == 0) ans = max(ans,j - i + 1);
}
}
cout<<ans<<'\n';
return 0;
}双指针（反转字符串、回文）
对撞指针
cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
string s;
cin>> s;
int flag = 1;
int n = s.length();
int l = 0, r = n - 1;
while(l < r){
if(s[l] == s[r]){
l++;
r--;
}
else {
flag = 0;
break;
}
}
if(flag) cout<<"Y";
else cout<<"N";
return 0;
}快慢指针
前缀和+快慢指针
cpp
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
int main(){
int n; // 序列长度
ll s; // 目标和阈值S
cin >> n >> s;
vector<ll> a(n); // 输入序列，大小为n
for(int i = 0; i < n; i++) cin >> a[i]; // 读入序列元素
// 构建前缀和数组，pf[i]表示前i个元素的和
vector<ll> pf(n+1, 0); // 初始化为0
for(int i = 0; i < n; i++){
pf[i+1] = pf[i] + a[i]; // pf[i+1] = a[0] + a[1] + ... + a[i]
}
int l = 0, r = 0; // 双指针，维护滑动窗口 [l, r)
int ans = n + 1; // 记录最短区间长度，初始化为不可能的值
while(r <= n){ // 当右指针未越界时继续
ll sum = pf[r] - pf[l]; // 计算区间 [l, r) 的和，即 a[l] + ... + a[r-1]
if(sum >= s){ // 如果当前区间和大于等于S
ans = min(ans, r - l); // 更新最短长度，区间长度为 r - l
l++; // 尝试缩短区间，左指针右移
}
else { // 如果当前区间和小于S
r++; // 扩大区间，右指针右移
}
}
if(ans == n + 1) cout << 0 << endl; // 如果ans未被更新，说明没有满足条件的区间
else cout << ans << endl; // 否则输出最短区间长度
return 0;
}滑动窗口
cpp
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+9;
int a[N];
int main(){
int n,s;
cin>>n>>s;
for(int i = 0; i< n;i++) cin>> a[i];
int l = 0,r = 0;
int w_sum = 0;
int ans = n + 1;
while(r<n){
w_sum += a[r];
while(w_sum >= s){
ans = min(ans,r - l + 1);
w_sum -= a[l];
l++;
}
r++;
}
cout<<(ans == n+1?0:ans);
return 0;
}