1.硬币兑换

#include
using namespace std;
int main()
{
int a[5000] = {0}, ans = 0;
for(int i = 1; i <= 2023; i++){
for(int j = i + 1; j <= 2023; j++){
a[i + j] += i;
ans = max(ans, a[i + j]);
}
}
cout<<ans;
return 0;
}
2.更小的数

#include
#include
using namespace std;
int main()
{
string s;
cin>>s;
int n = s.size(), ans = 0;
for(int i = 0; i < n; i++){
for(int j = i + 1; j < n; j++){
string t = s;
reverse(t.begin() + i, t.begin() + j + 1);
if(t < s) ans++;
}
}
cout<<ans;
return 0;
}
#include
using namespace std;
const int N = 5005;
int f[N][N], a[N];
int main()
{
string s;
cin>>s;
int n = s.size(), ans = 0;
for(int i = 1; i <= n; i++) a[i] = s[i - 1] - '0';
for(int len = 1; len < n; len++){
for(int i = 1, j = i + len; j <= n; i++, j = i + len){
if(a[i] == a[j]){
if(j - i == 1 || j - i == 2) f[i][j] = 0;
else f[i][j] = f[i + 1][j - 1];
}
if(a[i] > a[j]) f[i][j] = 1;
if(a[i] < a[j]) f[i][j] = 0;
if(f[i][j]) ans++;
}
}
cout<<ans;
return 0;
}
3.颜色平衡树

#include
#include
#include
using namespace std;
const int N = 2e5 + 10;
int h[N], e[N], ne[N], idx;
int n, c[N], ans;
unordered_map<int, int> mp;
void add(int x, int y){
e[idx] = y;
ne[idx] = h[x];
h[x] = idx;
idx++;
}
void dfs(int u, unordered_map<int, int> &m){
if(h[u] == -1){
4.买瓜

#include
#include
using namespace std;
double a[35], b[35];
int n, m, ans = 1e9, f;
bool cmp(const double &p, const double &q){
return p > q;
}
void dfs(int u, int cnt, double sum){
if(sum > m || sum + b[u] < m) return;
if(sum == m){
f = 1;
ans = min(ans, cnt);
return;
}
if(u > n) return;
dfs(u + 1, cnt, sum + a[u]);
dfs(u + 1, cnt + 1, sum + a[u] / 2);
dfs(u + 1, cnt, sum);
}
int main()
{
cin>>n>>m;
for(int i = 1; i <= n; i++) cin>>a[i];
sort(a + 1, a + n + 1, cmp);
for(int i = n; i > 0; i--) b[i] = b[i + 1] + a[i];
dfs(1, 0, 0);
if(!f) printf("-1");
else printf("%d", ans);
return 0;
}
5.网络稳定性

give up
6.异或和之和

#include
using namespace std;
const int N = 1e5 + 10;
int a[N], n;
long long ans;
int main()
{
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
for(int i = 0; i < n; i++){
int res = 0;
for(int j = i; j < n; j++){
res ^= a[j];
ans += res;
}
}
printf("%lld", ans);
return 0;
}
#include
using namespace std;
const int N = 1e5 + 10;
int a[N], cnt[25][5], n;
long long ans;
int main()
{
scanf("%d", &n);
for(int i = 1; i <= n; i++){
scanf("%d", &a[i]);
a[i] ^= a[i - 1];
}
for(int i = 0; i <= 20; i++)
for(int j = 0; j <= n; j++)
cnt[i][(a[j] >> i) & 1]++;
for(int i = 0; i <= 20; i++){
ans += (long long)cnt[i][0] * cnt[i][1] * (1 << i);
}
printf("%lld", ans);
}