A. Add Odd or Subtract Even
The meaning of problems: two to the number of xy, each of x plus or minus an odd an even, Q variable from x to y is the minimum number of operations.
Analysis: determine what parity x and y discuss it.
#include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int a,b;scanf("%d%d",&a,&b);
if(a>b){
printf("%d\n",((a&1)^(b&1))+1);
}
else if(a<b)printf("%d\n",(!((a&1)^(b&1)))+1);
else printf("0\n");
}
}B. WeirdSort
The meaning of problems: Given an array a, and an array of p, for any i, the array may exchange a number of types of pi and pi + 1 of the two positions. Q. Can a array into a non-descending order.
Analysis: determining a plurality of link blocks by an array of p, for each block Sort Unicom, then determines whether or not a ordered array.
#include <bits/stdc++.h>
using namespace std;
int main(){
int t;cin>>t;
while(t--){
int n,m;cin>>n>>m;
vector<int>a(n+1),p(m+1);
for (int i = 1; i <= n; ++i) {
scanf("%d",&a[i]);
}
for (int i = 1; i <= m; ++i) {
scanf("%d",&p[i]);
}
sort(p.begin()+1,p.end());
p.erase(unique(p.begin()+1,p.end()),p.end());
int fro=p[1],bak=p[1]+1;
for (int i = 2; i <= m; ++i) {
if(p[i]!=p[i-1]+1){
sort(a.begin()+fro,a.begin()+bak+1);
fro=p[i],bak=p[i]+1;
}
else bak=p[i]+1;
}
sort(a.begin()+fro,a.begin()+bak+1);
bool ok = 1;
for (int i = 2; i <= n; ++i) {
if(a[i]<a[i-1])ok=0;
}
if(ok)puts("YES");
else puts("NO");
}
}C. Perform the Combo
The meaning of problems: Given a string s and an array of p, for each pi, s traversed in 1-pi of letters, each letter Q traversal times.
Analysis: Differential.
#include <bits/stdc++.h>
using namespace std;
char s[200004];
int p[200004];
int a[200004];
int ans[26];
int main(){
int t;
cin>>t;
while(t--){
int n,m;scanf("%d%d",&n,&m);
scanf("%s",s+1);
for (int i = 0; i <= n; ++i) {
a[i]=0;
}
for (int i = 0; i < 26; ++i) {
ans[i]=0;
}
for (int i = 0; i < m; ++i) {
scanf("%d",&p[i]);
a[1]++,a[p[i]+1]--;
}
a[1]++;
int sum = 0;
for (int i = 1; i <= n; ++i) {
sum+=a[i];
ans[s[i]-'a']+=sum;
}
for (int i = 0; i < 26; ++i) {
printf("%d ",ans[i]);
}
puts("");
}
}D. Three Integers
The meaning of problems: a given number 3 ABC, each may be a -1 or +1 for any number, so that the operation can be asked several times happened A <= B <= C, B and A is a factor, B is a C factor.
Analysis: Violence enumeration, complexity is a harmonic progression.
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int a, b, c;
cin >> a >> b >> c;
int ans = 2e9,aa,bb,cc;
for (int i = 1; i <= 2e4; ++i) {
for (int j = i; j <= 2e4; j+=i) {
for (int k = j; k <= 2e4; k+=j) {
int temp = abs(a-i)+abs(b-j)+abs(c-k);
if(temp<ans){
ans = temp;
aa=i,bb=j,cc=k;
}
}
}
}
cout<<ans<<endl;
cout<<aa<<' '<<bb<<' '<<cc<<endl;
}
}E. Construct the Binary Tree
Meaning of the questions: Given n, d, n construct a point, the store d the depth of the tree and is.
Analysis: easy to think, the community and the depth of n points is a chain, the lower bound is a complete binary tree.
Then only need to calculate the number of points for each depth final answer can be obtained, a chain can be calculated as sum (n-1) * n / 2, then traversing the deepest points, each point of this put one up, sum-1, you only need to constantly put up the most points you can get a satisfying answer condition.
#include <bits/stdc++.h>
using namespace std;
int dep[5004];
int ans[5004];
vector<int>v[5004];
int main(){
int t;cin>>t;
while(t--){
int n,d;cin>>n>>d;
for (int i = 0; i <= n; ++i) {
v[i].clear();
}
int up = 0,dw = (n-1)*n/2;
int num = 0;
for (int i = 1; num <= n; i<<=1) {
num += i;
if(num <= n){
up += log2(i)*i;
}
else {
up += log2(i)*(i - num + n);
}
}
// printf("%d %d\n",up,dw);
if(d < up || d > dw){
puts("NO");
continue;
}
puts("YES");
memset(dep,0, sizeof(dep));
for (int i = 0; i < n; ++i) {
dep[i]=1;
}
int now = dw,top = 1,pos = n-1;
while(now > d){
int nxt = max(pos - now + d,top);
now -= pos - nxt;
dep[pos]--;
dep[nxt]++;
pos --;
if(dep[top] == dep[top-1]*2)top++;
}
int cnt = 2;
v[0].push_back(1);
for (int i = 1; dep[i] ; ++i) {
for (int j = 1; j <= dep[i]; ++j) {
ans[cnt]=v[i-1][(j-1)/2];
v[i].push_back(cnt++);
}
}
for (int i = 2; i <= n; ++i) {
printf("%d ",ans[i]);
}
puts("");
}
}F. Moving Points
The meaning of problems: n points, each point has a speed and a pos, D (U, V) is the minimum distance between two points UV, seeking sum D (U, V).
Analysis: for any two points, one point to another point greater than two attributes, its contribution to abs (pos1-pos2).
One dimension of the sort, another dimension bit sum contribution to maintaining statistics.
#include <bits/stdc++.h>
using namespace std;
struct BIT {
int n;
vector<long long> v;
BIT(int n) : n(n) { v.resize(n + 1); }
void update(long long x, long long d) {
while (x <= n) {
v[x] += d;
x += (x & -x);
}
}
long long que(long long x) {
long long res = 0;
while (x > 0) {
res += v[x];
x -= (x & -x);
}
return res;
}
};
BIT bit(200004),num(200004);
pair<int, int> p[200004];
unordered_map<int, int> mp;
int main() {
int n;
cin >> n;
vector<int> v;
for (int i = 1; i <= n; ++i) {
scanf("%d", &p[i].first);
}
for (int i = 1; i <= n; ++i) {
scanf("%d", &p[i].second);
v.push_back(p[i].second);
}
sort(v.begin(),v.end());
v.erase(unique(v.begin(), v.end()), v.end());
for (int i = 0; i < v.size(); ++i) {
mp[v[i]] = i + 1;
}
sort(p + 1, p + n + 1);
long long ans = 0;
for (int i = n; i >= 1; --i) {
ans += bit.que(mp[v[v.size()-1]]) -
bit.que(mp[p[i].second]-1) -
(num.que(mp[v[v.size()-1]]) - num.que(mp[p[i].second]-1))*(long long)(p[i].first - p[1].first);
bit.update(mp[p[i].second],p[i].first - p[1].first);
num.update(mp[p[i].second],1);
}
cout<<ans<<endl;
}
