A.1:41:18 solved by hl
大致就是一个图上染色判断是否有矛盾。
按照样例来看，似乎存在孤立点就要输出NO
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();}
while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1010;
const int maxm = 10010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,X,Y;
int fa[maxn],sz[maxn];
int color[maxn];
int head[maxn],tot;
struct Edge{
int to,next;
}edge[maxm * 2];
void init(){
for(int i = 0 ; i <= N ; i ++) fa[i] = i,sz[i] = 1;
for(int i = 0; i <= N ; i ++) head[i] = -1;
tot = 0;
}
int find(int x){
if(x == fa[x]) return x;
return fa[x] = find(fa[x]);
}
void Union(int a,int b){
a = find(a); b = find(b);
if(a != b){
sz[b] += sz[a];
fa[a] = b;
}
}
void add(int u,int v){
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
bool flag;
void dfs(int x){
//cout << x << " " << color[x] << endl;
for(int i = head[x]; ~i ;i = edge[i].next){
int v = edge[i].to;
if(!color[v]){
color[v] = -color[x];
dfs(v);
}
else if(color[v] == color[x]) flag = 0;
}
}
int main(){
while(~Sca2(N,M)){
Sca2(X,Y); init();
for(int i = 1; i <= N ; i ++) color[i] = 0;
for(int i = 1; i <= M ; i ++){
int u,v ; Sca2(u,v); add(u,v); add(v,u);
Union(u,v);
}
flag = 1;
for(int i = 1; i <= X ; i ++){
int x = read();
if(color[x] == -1) flag = 0;
if(!color[x]){
color[x] = 1;
dfs(x);
}
}
for(int i = 1; i <= Y; i ++){
int x = read();
if(color[x] == 1) flag = 0;
if(!color[x]){
color[x] = -1;
dfs(x);
}
}
for(int i = 1; i <= N ; i ++){
// cout <<sz[find(i)] << endl;
if(!color[i]){
if(sz[find(i)] == 1) flag = 0;
// if(flag) puts("YES");
// else puts("NO");
color[i] = 1;
dfs(i);
}
}
if(flag) puts("YES");
else puts("NO");
}
return 0;
}
A
C.4:34:31(-4) solved by zcz
方法1.如果较大的数和较小的数比例是黄金比就是0
方法2.设b为较大的数，a为较小的数，l为b - a,满足 b > b * l - a * a >= 0就为0
要用JAVA上大数
import java.math.*;
import java.util.*;
import java.io.*;
public class Main{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
BigInteger a,b,x,y,m,n,l,cnt,fu,q,zero;
while(in.hasNext()){
x = in.nextBigInteger();
y = in.nextBigInteger();
a = x; b = x;
b = b.max(y);
a = a.min(y);
l = b.subtract(a);
n = b.multiply(l);
q = a.multiply(a);
int t = b.compareTo(n.subtract(q));
int tt = n.compareTo(q);
if(t == -1 || tt == -1 || t == 0) System.out.println("1");
else System.out.println("0");
}
}
}
C
D. 1:14:59 solved by gbs
可以推公式，直接解方程解出来
x1= (X + sqrt(XX - 4ygcd(x,y) ) ) /2;
y1 = X - x1;
#include <iostream>
#include<stack>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<ctime>
#include<complex>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
using namespace std;
typedef long long LL;
int gcd(int a,int b)
{
if (b == 0)
return a;
return gcd(b,a%b);
}
int judge1(int X,int Y)
{
int left1 = 1;
int right1 = X/2;
while(left1<=right1)
{
//cout<<left1
int mid = (left1+right1)>>1;
LL ans = 1LL*(X-mid)*mid;
//cout<<ans<<‘ ‘<<Y<<‘ ‘<<mid<<endl;
if (ans>Y)
right1 = mid-1;
else if (ans == Y)
{
if (gcd(mid,X-mid)!=1)
return -1;
return mid;
}
else
left1 = mid+1;
}
return -1;
}
int main()
{
int X,Y;
int gcd1;
while(scanf("%d%d",&X,&Y)!=EOF){
gcd1 = gcd(X,Y);
X/=gcd1;
Y/=gcd1;
//cout<<X<<‘ ‘<<Y<<endl;
int ans1= judge1(X,Y);
if (ans1 == -1)
printf("No Solution\n");
else
printf("%d %d\n",ans1*gcd1,(X-ans1)*gcd1);
}
return 0;
}
/*
111
5111111 11111116 12133215 251111111
111
5 15 5 25
5 14 5 25
5 13 5 25
5 12 5 25
3
14 14 5 5
14 14 5 5
*/
D
F. 1:08:17(-1) solved by zcz and hl
WA:读错题
很显然，段数分的越多越好,同时要满足每段不同，所以从2开始分直到不能分
例如10分为 2 + 3 + 4,最后剩下的1平均从后往前分，变为 2 + 3 + 5
12分为 2 + 3 + 4 ,剩下的3分给他们变为 3 + 4 + 5
13分为 2 + 3 + 4,剩下的分为 3 + 4 + 5,又剩下1分到5上变为 3 + 4 + 6
这是一个等差数列，知道了分法就总结一下计算公式就可以了
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();}
while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N,M,K;
LL fac[maxn];
LL q_p(LL a,LL b){
LL ans = 1;
while(b){
if(b & 1) ans = ans * a %mod;
b >>= 1;
a = a * a % mod;
}
return ans;
}
LL inv(LL x){
return q_p(x,mod - 2);
}
void init(){
fac[0] = 1;
for(int i = 1; i < maxn; i ++) fac[i] = fac[i - 1] * i % mod;
}
LL check(LL x){
return (x + 2) * (x - 1) / 2;
}
LL solve(){
LL l = 2,r = INF;
LL ans = INF;
while(l <= r){
LL m = l + r >> 1;
if(N >= check(m)){
l = m + 1;
ans = m;
}else{
r = m - 1;
}
}
return ans;
}
int main(){
int T; Sca(T); init();
LL inv2 = inv(2);
while(T--){
Sca(N);
if(N <= 4){
Pri(N);
continue;
}
LL n = solve();
LL a = N - check(n);
// cout << a << " " << n <<endl;
LL ans = fac[n + 1] * inv(n + 1 - a) % mod;
if(n == a){
ans = (n + 2) * fac[n] % mod *inv2 % mod;
}
Prl(ans);
}
return 0;
}
F
G.3:05:25(-2) solved by hl
WA:没整清楚就交 + 忘记特判
点分治 + 状压dp
点分治每次寻找子树的重心，对于不同的两棵树之间维护dp1[i]表示已经被记入的树在i满足存在状态i结点的数量
dp2表示正在处理的当前子树的状态数，dp3[i]表示当前处理的子树下状态确实为i的数量
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();}
while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;}
const double PI = acos(-1.0);
const double eps = 1e-9;
const int maxn = 5e4 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
LL N,M,K;
int type[maxn];
struct Edge{
int to,next;
}edge[maxn * 2];
int head[maxn],tot;
void init(){
for(int i = 0 ; i <= N ; i ++) head[i] = -1;
tot = 0;
}
void add(int u,int v){
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
int root,max_part,SUM,ed;
int size[maxn],vis[maxn];
void dfs_root(int u,int la){
size[u] = 1;
int heavy = 0;
for(int i = head[u];~i ; i = edge[i].next){
int v = edge[i].to;
if(v == la || vis[v]) continue;
dfs_root(v,u);
heavy = max(heavy,size[v]);
size[u] += size[v];
}
if(max_part > max(heavy,SUM - heavy)){
max_part = max(heavy,SUM - heavy);
root = u;
}
}
LL dp1[3000],dp2[3000],dp3[3000];
void dfs_dis(int t,int la,int state){
state |= (1 << type[t]);
dp3[state]++;
for(int sta = state;sta; sta = (sta - 1) & state) dp2[sta]++; dp2[0]++;
for(int i = head[t]; ~i ; i = edge[i].next){
int v = edge[i].to;
if(vis[v] || v == la) continue;
dfs_dis(v,t,state);
}
}
LL ans;
void work(int t){
int state = 1 << type[t];
for(int i = 0 ; i <= ed; i ++) dp1[i] = dp2[i] = dp3[i] = 0;
for(int sta = state;sta; sta = (sta - 1) & state) dp1[sta]++; dp1[0]++;
LL sum = 0;
for(int i = head[t]; ~i ; i = edge[i].next){
int v = edge[i].to;
if(vis[v]) continue;
dfs_dis(v,t,state);
for(int sta = 0; sta <= ed; sta++){
ans += dp1[sta] * dp3[ed - sta];
// cout << " " << sta <<" " << dp1[sta] << " " << dp3[ed - sta] << endl;
}
//ans -= sum;
// sum += dp2[ed - state];
for(int sta = 0; sta <= ed; sta++){
dp1[sta] += dp2[sta];
dp2[sta] = 0;
dp3[sta] = 0;
}
// cout << t << " " << v << " " << ans << endl;
}
}
void divide(int t){
max_part = INF,root = t;
dfs_root(t,-1);
vis[root] = 1;
work(root);
for(int i = head[root]; ~i ; i = edge[i].next){
int v = edge[i].to;
if(vis[v]) continue;
SUM = size[v];
divide(v);
}
}
int main(){
while(~scanf("%lld%d",&N,&K)){
init(); ed = (1 << K) - 1; ans = 0;
for(int i = 0 ; i <= N ; i ++) vis[i] = 0;
for(int i = 1; i <= N ; i ++){
Sca(type[i]);
type[i]--;
}
for(int i = 1; i <= N - 1; i ++){
int u,v; Sca2(u,v); add(u,v); add(v,u);
}
if(K == 1){
Prl(N * N);
continue;
}
SUM = N; divide(1);
Prl(ans *2);
}
return 0;
}
/*
3 2
1 2 2
1 2
1 3
0 0 1
1 1 0
2 0 0
3 0 0
1 3 0
0 0 1
1 2 0
2 1 0
3 1 0
1 2 0
0
*/
G
H.0:27:19 solved by gbs
判断一下奇偶
#include <iostream>
#include<stack>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<ctime>
#include<complex>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
using namespace std;
typedef long long LL;
int main()
{
int n;
while(scanf("%d",&n)!=EOF){
if (n&1)
printf("0\n");
else
printf("1\n");
}
return 0;
}
/*
111
5111111 11111116 12133215 251111111
111
5 15 5 25
5 14 5 25
5 13 5 25
5 12 5 25
3
14 14 5 5
14 14 5 5
*/
H
I.0:16:52 solved by hl
把凸包变成一圈三角形拼起来的，
三角形面积公式是 S = 0.5absin(θ)
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <bitset>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<‘0‘ || c>‘9‘){if (c == ‘-‘) f = -1;c = getchar();}
while (c >= ‘0‘&&c <= ‘9‘){x = x * 10 + c - ‘0‘;c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 110;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int N;
double M,K;
double a[maxn];
const double PI = acos(-1.0);
int main(){
while(~Sca(N)){
scanf("%lf",&M);
for(int i = 1; i <= N; i ++){
scanf("%lf",&a[i]);
}
double ans = 0;
for(int i = 1; i <= N ; i ++){
ans += M * M / 2 * sin(a[i] * PI/ 180);
}
printf("%.3lf\n",ans);
}
return 0;
}
I
J. 0:10:46 solved by gbs
每次八位判断一下是不是97
#include <iostream>
#include<stack>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<ctime>
#include<complex>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
using namespace std;
typedef long long LL;
int main()
{
int n;
unsigned int h1;
while(cin >>n)
{
int ans =0;
for (int i=0; i<n; i++)
{
scanf("%u",&h1);
while(h1)
{
//cout<<(h1&255)<<endl;
if ( (h1&255) ==97 )
ans++;
h1>>=8;
}
}
printf("%d\n",ans);
}
}
/*
111
5111111 11111116 12133215 251111111
111
5 15 5 25
5 14 5 25
5 13 5 25
5 12 5 25
3
14 14 5 5
14 14 5 5
*/
J
K. 3:15:49(-1) solved by zcz
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn=5e6+5;
long long mod=100000073;
long long a[maxn];
long long b[maxn];
long long sum[maxn];
int main()
{
int c1=1,c2=1;
while(c2<maxn)
{
a[c2]=c1;
c1++;
c2+=c1;
}
a[maxn-1]=c1;
for(int i=maxn-2;i>=1;i--)
{
if(!a[i]) a[i]=a[i+1];
}
b[1]=1;
b[2]=2;
b[3]=1;
sum[1]=1;
sum[2]=3;
sum[3]=4;
c1=0;
for(int i=4;i<maxn;i++)
{
if(c1==0) c1=a[i];
//cout<<c1<<endl;
// for(int j=0,k=a[i]-1;j<c1;j++,k--)
// {
// b[i]=(b[i]+b[i-k-1])%mod;
// }
b[i]=(sum[i-a[i]+c1-1]-sum[i-a[i]-1]+mod)%mod;
sum[i]=(sum[i-1]+b[i])%mod;
c1--;
}
long long a1,a2;
while(scanf("%lld %lld",&a1,&a2)!=EOF)
{
printf("%lld %lld\n",a[a2-a1+1],b[a2-a1+1]);
}
return 0;
}
K
原文地址：https://www.cnblogs.com/Hugh-Locke/p/11328979.html
时间： 2024-11-09 01:47:29