A Wrestling Match
判断二分图，特判没属性的孤点。
#include <bits/stdc++.h>
using namespace std;
const int maxn=1050;
vector<int> e[maxn];
int n,m,x,y;
int colour[maxn];
int du[maxn];
int vis[maxn];
bool dfs(int v,int c) {
colour[v]=c;
for(int i=0;i<(int)e[v].size();i++) {
if(colour[e[v][i]]==c) {
return false;
}
if(colour[e[v][i]]==0&&!dfs(e[v][i],-c))
return false;
}
return true;
}
int main(){
while(scanf("%d%d%d%d",&n,&m,&x,&y)!=EOF){
for(int i=0;i<=n;i++) e[i].clear();
memset(du,0,sizeof du);
while(m--){
int u,v;
scanf("%d%d",&u,&v);
e[u].push_back(v);
e[v].push_back(u);
du[u]++;du[v]++;
}
memset(colour,0,sizeof(colour));
memset(vis,0,sizeof vis);
bool ptd=0;
for(int i=1;i<=x;i++){
int x;
scanf("%d",&x);
colour[x]=1;
vis[x]=1;
}
for(int i=1;i<=y;i++){
int x;
scanf("%d",&x);
colour[x]=-1;
if(vis[x]==1){
ptd=1;
break;
}
vis[x]=-1;
}
if(ptd){
puts("NO");
continue;
}
for(int i=1;i<=n;i++){
if(vis[i]==0 && du[i]==0){
puts("NO");
ptd=1;
break;
}
}
if(ptd)continue;
bool flag=true;
for(int i=1;i<=n && flag;i++){
if(colour[i]==1) flag=dfs(i,1);
else if(colour[i]==-1) flag=dfs(i,-1);
}
for(int i=1;i<=n && flag;i++){
if(colour[i]==0) flag=dfs(i,1);
}
// for(int i=1;i<=n;i++)
// printf("%d ",colour[i]);
// puts("");
for(int i=1;i<=n;i++){
if(colour[i]==0){
flag=false;
break;
}
}
if(flag) puts("YES");
else puts("NO");
}
return 0;
}
C Game of Taking Stones
威佐夫博弈
求$frac{sqrt(5)+1}{2}b$。
import java.util.*;
import java.math.*;
public class Main{
public static void main(String[] args){
Scanner cin = new Scanner(System.in);
BigDecimal k = null;
BigDecimal eps = BigDecimal.valueOf(0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001);
BigDecimal l=BigDecimal.valueOf(1.0);
BigDecimal r=BigDecimal.valueOf(3.0);
BigDecimal Five=BigDecimal.valueOf(5.0);
for(int i=1;i<=2000;i++){
BigDecimal mid=l.add(r);
mid=mid.divide(BigDecimal.valueOf(2.0));
BigDecimal p = mid.multiply(mid);
int pk = p.compareTo(Five);
if(pk<=0){
l=mid.add(eps);
k=mid;
} else {
r=mid.subtract(eps);
}
}
//System.out.println(k);
k=k.add(BigDecimal.valueOf(1.0));
k=k.divide(BigDecimal.valueOf(2.0));
BigDecimal a,b;
while(cin.hasNext()){
a=cin.nextBigDecimal();b=cin.nextBigDecimal();
int pk=a.compareTo(b);
if(pk<0){
BigDecimal tmp = a;
a=b;
b=tmp;
}
BigDecimal ans = a.subtract(b);
ans=ans.multiply(k);
ans=ans.subtract(b);
pk = ans.compareTo(BigDecimal.valueOf(1.0));
int pt=ans.compareTo(BigDecimal.valueOf(0.0));
if(pk<0 && pt>=0) System.out.println(0);
else System.out.println(1);
}
}
}
D A Simple Math Problem
设$G=gcd(a,b)$。
$gcd(a,b)=gcd(x+y,lcm(x,y))=gcd(x+y,ky)=gcd(x+y,y)=gcd(x,y)$
那么等式$frac{x}{G}+frac{y}{G}=frac{a}{G}$。
$lcm(x,y)=frac{xy}{gcd(x,y)}$,
$frac{x}{G}frac{y}{G}=frac{b}{G}$.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a % b);
}
ll a, b, x, y;
ll getsqrt(ll a){
if(a==0) return 0;
ll l=1,r=1e9;
while(l<=r){
ll mid = (l+r)/2;
ll p=mid*mid;
if(p==a)
return mid;
if(p<a)
l=mid+1;
if(p>a)
r=mid-1;
}
return -1;
}
int main() {
//freopen("in.txt","r",stdin);
while (~scanf("%lld%lld", &a, &b)) {
ll G = gcd(a, b);
ll aa = a / G;
ll bb = b / G;
ll d = aa * aa - 4ll * bb;
if(d<0){
puts("No Solution");
continue;
}
ll t = getsqrt(d);
//cout<<t<<endl;
if(t==-1){
puts("No Solution");
continue;
}
ll ans1=(aa-t)/2;
ll ans2=(aa+t)/2;
if(ans1*2ll!=aa-t){
puts("No Solution");
continue;
}
ans1*=G;
ans2*=G;
if(ans1>ans2) swap(ans1,ans2);
if(ans1<=0 || ans2>a){
puts("No Solution");
continue;
}
printf("%lld %lld
",ans1,ans2);
}
return 0;
}
F Detachment
连续整数相乘时值很大。考虑到1没有贡献，那么必然是从2开始。
5=2+3
9=2+3+4
14=2+3+4+5
....
这些是分界。
多余的时候挂在末尾，都加1就行。
预处理fac和inv。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 1e5 + 5;
const int N = 1e5;
ll fac[maxn];
ll inv[maxn];
ll S[maxn];
inline ll power(ll a, ll n, ll p) {
ll ret = 1; ll now = a;
while (n != 0) {
if (n & 1)
ret = ret * now % p;
now = now * now % p;
n >>= 1;
}
return ret;
}
void init() {
fac[0] = 1;
for (ll i = 1; i <= N; i++)
fac[i] = fac[i - 1] * i % mod;
inv[N] = power(fac[N], mod - 2, mod);
for (ll i = N; i >= 1; i--)
inv[i - 1] = inv[i] * i % mod;
S[2]=2;
for(ll i=3;i<=N;i++)
S[i]=S[i-1]+i;
}
int main() {
init();
int t;
ll n;
scanf("%d", &t);
while (t--) {
scanf("%lld", &n);
if (n <= 4) {
printf("%lld
", n);
continue;
}
ll x;
ll l = 2, r = N;
while (l <= r) {
ll mid = (l + r) / 2;
if ( S[mid] <= n) {
l = mid + 1;
x = mid;
} else {
r = mid - 1;
}
}
x--;
ll avg = n / x;
ll ans = 1;
if (x & 1) {
ll k = x / 2;
ll left = avg - k;
ll right = avg + k;
ll sum = (left + right) * x / 2;
ll tmp = n - sum;
while (tmp >= x) {
tmp -= x;
left++;
right++;
}
if (tmp == 0) {
ans = fac[right] * inv[left - 1] % mod;
} else {
ll lright = right - tmp;
right++;
ans = fac[lright] * inv[left - 1] % mod * fac[right] % mod * inv[lright + 1] % mod;
}
} else {
ll k = x / 2;
ll left = avg - k;
ll right = avg + k - 1;
ll sum = (left + right) * x / 2;
ll tmp = n - sum;
while (tmp >= x) {
left++; right++;
tmp -= x;
}
if (tmp == 0) {
ans = fac[right] * inv[left - 1] % mod;
} else {
ll lright = right - tmp;
right++;
ans = fac[lright] * inv[left - 1] % mod * fac[right] % mod * inv[lright + 1] % mod;
}
}
printf("%lld
", ans % mod);
}
return 0;
}
H To begin or not to begin
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
//freopen("in.txt","r",stdin);
int n;
while(~scanf("%d",&n))
printf("%d
",(n+1)%2);
return 0;
}
I Convex
#include <bits/stdc++.h>
using namespace std;
const int maxn=20;
const double pi=acos(-1);
int a[1111];
int main() {
int n,d;
while(scanf("%d%d",&n,&d)!=EOF) {
for(int i=1;i<=n;i++) {
scanf("%d",&a[i]);
}
double ans = 0.0;
for(int i=1;i<=n;i++) {
ans+=0.5*d*d*sin(a[i]*pi/180.0);
}
printf("%.3lf
",ans);
}
return 0;
}
J Find Small A
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
//freopen("in.txt","r",stdin);
int x,n;
while(~scanf("%d",&n)){
int cnt = 0;
for(int i=1;i<=n;i++){
scanf("%d",&x);
while(x){
if(x % 256 == 97)
cnt++;
x/=256;
}
}
cout<<cnt<<endl;
}
return 0;
}