AtCoder Beginner Contest 190
A - Very Very Primitive Game
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int inf=1e18;
const int mod = 1e9 + 7;
const int maxn = 3e5 + 10;
int a[maxn];
void solve() {
int a,b,c;
cin>>a>>b>>c;
if (a==b){
cout<<(c==1?"Takahashi":"Aoki");
} else{
cout<<(a>b?"Takahashi":"Aoki");
}
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
// cin >> _;
while (_--) {
solve();
}
return 0;
}
B - Magic 3
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int inf=1e18;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int a[maxn];
int b[maxn];
void solve() {
int n,s,d;
cin>>n>>s>>d;
for (int i = 0; i < n; ++i) {
cin>>a[i]>>b[i];
}
int f=0;
for (int i = 0; i < n; ++i) {
if (a[i]<s&&b[i]>d) {
f=1;
}
}
cout<<(f==1?"Yes\n":"No\n");
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
// cin >> _;
while (_--) {
solve();
}
return 0;
}
C - Bowls and Dishes
二进制硬暴力每种情况
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int inf=1e18;
const int mod = 1e9 + 7;
const int maxn = 1e2 + 10;
int a[maxn];
int b[maxn];
int c[maxn];
int d[maxn];
int e[maxn];
void solve() {
int n,m;
cin>>n>>m;
for (int i = 0; i < m; ++i) {
cin>>a[i]>>b[i];
}
int k;
cin>>k;
for (int i = 0; i < k; ++i) {
cin>>c[i]>>d[i];
}
int ans=0;
for (int i = 0; i < (1 << k); ++i) {
memset(e,0,sizeof(e));
for (int j = 0; j <k; ++j) {
//cout<<((i>>j)&1)<<"\n";
if ((i>>j)&1){
e[c[j]]=1;
} else e[d[j]]=1;
}
/*for (int l = 1; l <=m; ++l) {
cout<<e[l]<<" ";
}cout<<"\n";*/
int sum=0;
for (int j = 0; j < m; ++j) {
//cout<<e[a[j]]<<" "<<e[b[j]]<<"\n";
if (e[a[j]]==1&&e[b[j]]==1){
sum++;
}
}//cout<<"\n\n";
ans=max(ans,sum);
}
cout<<ans;
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
// cin >> _;
while (_--) {
solve();
}
return 0;
}
D - Staircase Sequences
求其奇数因子个数，因为可以配合负数所以每种都可以变成俩种，所以ans=2*奇数因子个数
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int inf = 1e18;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
int dp[maxn];
void solve() {
int a ;
cin >> a;
int ans = 0; //ans是求a所有因子数的总个数
for (int i = 1; i * i <= a; i++) {
if (a % i == 0) {
if (i%2==1) ans++;
if (i != a / i&&(a/i)%2==1) {
//两个因子数不相等
ans++;
}
}
}
printf("%d\n", 2*ans);
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
// cin >> _;
while (_--) {
solve();
}
return 0;
}
F - Shift and Inversions
先求出原数组逆序对，然后每把开头移到结尾，ans+=(n-b[i]-1)-b[i];
因为把b[i]放到结尾前面有n-b[i]-1比他大，又原本有b[i]个逆序因为他的移到而没有了
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int inf = 1e18;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
int a[maxn];
int t[maxn];
int b[maxn];
int res;
void merge_sort(int l,int r){
if(l>=r)return ;
int mid=l+r>>1;
merge_sort(l,mid),merge_sort(mid+1,r);//递归分治
int i=l,j=mid+1,n=0;
while(i<=mid&&j<=r)
if(a[i]<=a[j])t[n++]=a[i++];
else res+=mid-i+1,t[n++]=a[j++];//计算答案的情况
while(i<=mid)t[n++]=a[i++];
while(j<=r)t[n++]=a[j++];
for(int i=l,j=0;i<=r;i++,j++)a[i]=t[j];
}
void solve() {
int n;
cin>>n;
for (int i = 0; i < n; ++i) {
cin>>a[i],b[i]=a[i];
}
merge_sort(0,n-1);
for (int i = 0; i < n; ++i) {
cout<<res<<"\n";//cout<<(n-b[i]-1)<<"\n";
res+=(n-b[i]-1)-b[i];
}
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
// cin >> _;
while (_--) {
solve();
}
return 0;
}
//0 3 1 5 4 2 9 6 8 7 9
//3 1 5 4 2 9 6 8 7 0 9+(10-0-1)=18
//1 5 4 2 9 6 8 7 0 3 18+(10-3-1)=24
//5 4 2 9 6 8 7 0 3 1
//4 2 9 6 8 7 0 3 1 5