A题：
#include <bits/stdc++.h>
using namespace std;
const int N = 50;
int a[N];
int main()
{
int t;
cin >> t;
while(t --)
{
int n;
cin >> n;
for(int i = 0;i < n;i ++) cin >> a[i];
int cnt = 0;
for(int i = 0;i < n - 1;i ++)
{
// cout << i << ' ' << a[i] << ' ' << cnt << endl;
if(a[i + 1] >= a[i]){
if((a[i + 1] + a[i] - 1) / a[i] <= 2) continue;
a[i] *= 2;
cnt ++;
i --;
}else{
if((a[i] + a[i + 1] - 1) / a[i + 1] <= 2) continue;
// a[i] /= 2;
if(a[i] & 1){
a[i] = a[i] / 2 + 1;
}else{
a[i] = a[i] / 2;
}
cnt ++;
i --;
}
}
cout << cnt << endl;
}
return 0;
}
B题：
- 让余数等于0的等于余数为1，并且等于余数为2的。其实就只需要统计出等于0的数字个数，等于1的数字个数，等于2的数字个数。如果0的多余，那么就转移给1，如果1的多余转移给2，如果2多余，就转移给0
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int a[N];
int main()
{
int t;
cin >> t;
while(t --)
{
int n;
cin >> n;
a[0] = 0,a[1] = 0,a[2] = 0;
for(int i = 0,x;i < n;i ++){
cin >> x;
a[x % 3] ++;
}
int ave = n / 3;
int cnt = 0;
for(int i = 0;i < 10;i ++){
if(a[2] > ave){
cnt += a[2] - ave;
a[0] += a[2] - ave;
a[2] = ave;
}
if(a[0] > ave){
cnt += a[0] - ave;
a[1] += a[0] - ave;
a[0] = ave;
}
if(a[1] > ave){
cnt += a[1] - ave;
a[2] += a[1] - ave;
a[1] = ave;
}
}
cout << cnt << endl;
}
return 0;
}
C题：
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
typedef long long LL;
int main()
{
int t;
cin >> t;
unordered_set<LL> st;
for(LL i = 1;i <= 10000;i ++){
st.insert(i * i * i);
}
while(t --)
{
LL n;
cin >> n;
bool flag = false;
for(LL i = 1;i <= 10000;i ++){
LL b = n - i * i * i;
if(st.count(b)){
flag = true;
}
}
if(flag){
cout << "YES" << endl;
}else{
cout << "NO" << endl;
}
}
return 0;
}
D题：
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int t,n;
int a[N];
int deep[N];
void build(int l,int r,int dep)
{
if(l == r){
return;
}
int ml = l;
for(int i = l;i < r;i ++){
if(a[i] > a[ml]){
ml = i;
}
}
// cout << ml << ' ' << a[ml] << endl;
deep[a[ml]] = dep + 1;
build(l,ml,dep + 1);
build(ml + 1,r,dep + 1);
}
int main()
{
cin >> t;
while(t --)
{
memset(deep,0,sizeof deep);
cin >> n;
int maxn = 1;
for(int i = 1;i <= n;i ++)
{
cin >> a[i];
if(a[i] >= a[maxn]){
maxn = i;
}
}
deep[maxn] = 0;
build(1,maxn,0);
build(maxn + 1,n + 1,0);
for(int i = 1;i <= n;i ++){
cout << deep[a[i]] << ' ';
}
cout << endl;
}
return 0;
}
E题
- 二分题，大于等于的就可以把比他小的吃掉，而且还能获得他拥有的物品
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int t,n;
typedef long long LL;
LL sum[N];
int a[N];
int b[N];
bool check(int x)
{
long long ans = sum[x];
for(int i = x + 1;i <= n;i ++){
if(ans >= a[i]){
ans += a[i];
}else{
return false;
}
}
return true;
}
int main()
{
cin >> t;
while(t --)
{
cin >> n;
for(int i = 1;i <= n;i ++){
cin >> a[i];
b[i] = a[i];
}
sort(a + 1,a + n + 1);
for(int i = 1;i <= n;i ++){
sum[i] = sum[i - 1] + a[i];
// b[i] = a[i];
}
int l = 1,r = n;
while(l < r)
{
int mid = l + r >> 1;
if(check(mid)){
r = mid;
}else{
l = mid + 1;
}
}
cout << n - l + 1 << endl;
for(int i = 1;i <= n;i ++){
if(b[i] >= sum[l] - sum[l - 1])
cout << i << ' ';
}
cout << endl;
}
return 0;
}
F题
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1e6 + 10;
int a[N];
int sum[N];
int main()
{
ios::sync_with_stdio(false);
int t;
cin >> t;
while(t -- )
{
int n;
cin >> n;
map<int,int>mp;
for(int i = 0,x;i < n;i ++){
cin >> x;
mp[x] ++;
a[i] = 0;
}
a[n] = 0;
for(auto &x : mp){
a[x.second] ++;
}
vector<pair<int,int>> v;
int ans = n;
for(int i = 1;i <= n;i ++){
if(a[i] == 0) continue;
v.push_back({i,a[i]});
}
sum[0] = v[0].second;
for(int i = 1;i < v.size();i ++){
sum[i] = sum[i - 1] + v[i].second;
}
int sum1 = 0;
for(int i = 0;i < v.size();i ++){
int t1 = (sum[v.size() - 1] - sum[i]) * v[i].first;
int t2 = n - sum1 - v[i].second * v[i].first - t1;
ans = min(ans,t2 + sum1);
// cout << ans << endl;
sum1 += v[i].second * v[i].first;
}
cout << ans << endl;
}
return 0;
}
/*
1 1
2 2
3 1
*/
G题
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL N = 1e6 + 10;
LL a[N];
LL x[N];
LL sum[N];
LL b[N];
LL t,n,m;
vector<pair<LL,LL>> v;
LL mmax;
long long dfs(LL i)
{
int cur = x[i];
if(sum[n] <= 0 && mmax < x[i]){
return -1;
}
LL f1,f2 = 0;
if(mmax >= x[i]){
f1 = 0;
f2 = x[i];
}else{
f1 = (x[i]-mmax-1)/sum[n]+1; // 这个地方是个重点，要下取整，而且同时避免是倍数的时候
f2 = x[i] - sum[n] * f1;
}
LL l = 0,r = n - 1;
while(l < r){
LL mid = l + r >> 1;
if(v[mid].second < f2){
l = mid + 1;
}else{
r = mid;
}
}
return f1 * n + v[l].first - 1;
}
int main()
{
cin >> t;
while(t --)
{
cin >> n >> m;
v.clear();
for(LL i = 1;i <= n;i ++)
cin >> a[i];
for(LL i = 1;i <= m;i ++)
cin >> x[i];
map<LL,LL>mp;
mmax = -1e10;
for(LL i = 1;i <= n;i ++){
sum[i] = sum[i - 1] + a[i];
mmax = max(sum[i],mmax);
v.push_back({i,mmax});
}
sort(v.begin(),v.end(),[](pair<LL,LL>f1,pair<LL,LL>f2){
if(f1.second == f2.second){
return f1.first < f2.first;
}else{
return f1.second < f2.second;
}
});
for(LL i = 1;i <= m;i ++){
cout << dfs(i) << ' ';
}
cout << endl;
}
}