序
题号 标题 已通过代码 通过率 团队的状态
A 小凯的疑惑 点击查看 434/745 通过
B 幻象踩花 点击查看 121/434 通过
C 跳刀抓人 点击查看 26/178 通过
D 真正的卡尔 点击查看 115/714 通过
E 刷新的艺术 点击查看 31/479 通过
F 最后的战役 点击查看 1/39 未通过
G 随机数生成器 点击查看 1/20 未通过
H 字符串 点击查看 25/130 通过
I 修罗牌浪 点击查看 1/4 未通过
J 集合论 点击查看 72/242 通过
K 有趣的矩阵 点击查看 6/20 未通过
A、小凯的疑惑
- NOIP2017D1T1原题，小学奥数公式，结论是ab-a-b。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
LL ksm(LL a, LL b){if(b==0)return 1;LL ns=ksm(a,b>>1);ns=ns*ns%mod;if(b&1)ns=ns*a%mod;return ns;}
void EXGCD(int a, int b, int &d, int &x, int & y, int MOD) { if (b==0) { d = a; x = 1; y = 0; } else { EXGCD(b, a % b, d, y, x, MOD); y -= x * (a / b); } }
int inv(int a, int MOD) { int d=0, x=0, y=0; EXGCD(a, MOD, d, x, y, MOD); return d == 1 ? (x + MOD) % MOD : -1; }
int main(){
LL a, b; cin>>a>>b;
cout<<a*b-a-b;
return 0;
}B、幻象踩花
- 画图以后结论题，注意特判
- 开始思路错了几个方向，后来特判想了好一会儿，
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double pi = acos(-1);
int main(){
//cout<<area(point{2,0},1, point{-2,-2},2.8)<<"\n";
int T; cin>>T;
while(T--){
LL R, r, x, y; cin>>R>>r>>x>>y;
if(x*x+y*y>(R+r)*(R+r) || x*x+y*y<=(R-r)*(R-r)){
printf("%.10lf\n",0 ); continue;
}
double d = sqrt(x*x+y*y);
double t = (x*x+y*y+R*R-r*r)*1.0/(2*d*R);
//cout<<t<<"\n";
double ans = acos(t)/pi*2;
printf("%.10lf\n",ans );
}
return 0;
}C、跳刀抓人
- 暴力bfs，对于闪烁操作，在每个节点中多维护一个time表示当前是第几秒，转移时特判当前能否转移。
- 没有判断能否转移，直接全部暴力转移了，WA了好久
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double pi = acos(-1);
struct node{int x, y, t, time;};
struct cmp{
bool operator()(node a, node b){ return a.t > b.t; }
};
priority_queue<node, vector<node>, cmp>q;
int xx[] = {0,0,-1,1};
int yy[] = {1,-1,0,0};
char a[110][110];
int vis[110][110][10];
int main(){
int n, m; cin>>n>>m;
node st, ed;
for(int i = 1; i <= n; i++){
//scanf("%s",a[i]+1);
for(int j = 1; j <= m; j++){
cin>>a[i][j];
if(a[i][j]=='A')st = node{i,j,0,7};
if(a[i][j]=='B')ed = node{i,j,0,7};
}
}
// for(int i = 1; i <= n; i++){
// for(int j = 1; j <= m; j++){
// cout<<a[i][j];
// }
// cout<<"\n";
// }
//queue<node>q;
q.push(st);
while (q.size()){
node t = q.top(); q.pop();
t.time = min(t.time, 7);
if(vis[t.x][t.y][t.time])continue;
vis[t.x][t.y][min(t.time,7)] = 1;
if(a[t.x][t.y]=='B'){
cout<<t.t<<"\n";
return 0;
}
for(int dx = -3; dx <= 3; dx++){
for(int dy=-3; dy<=3; dy++){
if(dx==0&&dy==0)continue;
int nx=t.x+dx, ny = t.y+dy;
if(nx<1||nx>n||ny<1||ny>m)continue;
if(a[nx][ny]=='#')continue;
if(t.time==7)q.push({nx,ny,t.t+1,0});
// int nx = t.x+dx, ny = t.y+dy;
// if(nx<1||nx>n||ny<1||ny>m)continue;
// if(a[nx][ny]=='#'|| vis[t.x][t.y][nx][ny])continue;
// int dd = abs(dx)+abs(dy), nt;
// if(dd==1)nt = 1;else nt=8;
// if(ok==0){nt=1;}
// q.push(node{nx,ny,t.t+nt});
// vis[t.x][t.y][nx][ny] = 1;
}
}
int d = 7-t.time;
if(t.time<7)q.push({t.x,t.y,t.t+d,7});
for(int i = 0; i < 4; i++){
int nx = t.x+xx[i], ny = t.y+yy[i];
if(nx<1||nx>n||ny<1||ny>m)continue;
if(a[nx][ny]=='#')continue;
q.push({nx,ny,t.t + 1,t.time + 1});
}
}
cout<<-1<<'\n';
return 0;
}D、真正的卡尔
- 排列组合结论题，打表找规律得到递推公式
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod = 998244353;
const int maxn = 1010;
LL f[maxn][maxn];
int main(){
for(int i = 1; i <= 1000; i++)f[i][1] = i, f[1][i] = 1;
for(int i = 2; i <= 1000; i++){
for(int j = 2; j <= 1000; j++){
f[i][j] = (f[i-1][j]+f[i][j-1])%mod;
}
}
int T; cin>>T;
while(T--){
int a, b; cin>>a>>b;
cout<<f[a][b]<<"\n";
}
return 0;
}E、刷新的艺术
- 分类讨论结论题，开始有两种情况没有想到
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double pi = acos(-1);
int gcd(int a, int b){ return b==0 ? a : gcd(b,a%b); }
void out(int x, int y){
if(x%y==0)cout<<x/y<<"\n";
else cout<<x/gcd(x,y)<<"/"<<y/gcd(x,y)<<"\n";
}
int main(){
int T; cin>>T;
while(T--){
int a, b; cin>>a>>b;
if(a>b){
//cout<<b<<"\n"; continue;
if(a>b*2)cout<<b<<"\n";else out(a,2);
continue;
}
//out(b, b/a+1);
if(b%a==0)out(b,b/a+1);
else{
int c = b/a+1;
if(b*1.0/c < a*1.0*c/(c+1))out(b,c);
else out(a*c,c+1);
}
}
return 0;
}H、字符串
- KMP暴力出所有区间，然后贪心，按照左端点排序扫一遍更新答案
- 扫的时候需要考虑区间右边的重合情况，额外开个队列维护
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e7+100;
int n, m, k; string s,p;
struct node{int l, r;};
bool cmp(node x, node y){
return x.l<y.l;
}
vector<node>vc;
int nxt[maxn];
void pre_next(string p){
int j = 0; //j表示到i-1为止的最长前缀的下标
for(int i = 2; i < p.size(); i++){
//用到j为止的最长公共前缀的下标，尝试去匹配下一个字母
while(j && p[i]!=p[j+1])j = nxt[j];
if(p[i]==p[j+1])j++;
nxt[i] = j;
}
}
void kmp(string s, string p){
int j = 0;
for(int i = 1; i <= s.size()-1; i++){
while(j && s[i]!=p[j+1])j = nxt[j];//不停的去找直到s[i]能匹配上下一个
if(s[i]==p[j+1])j++;//匹配上了就加一
if(j==p.size()-1){//匹配完了
int t = i-p.size()+1;
//cout<<t+1<<" ";
vc.push_back(node{t+1,t+m});
j = nxt[j];
}
}
}
int main(){
cin>>n>>m>>k>>s>>p;
s="0"+s; p="0"+p;
pre_next(p);
kmp(s,p);
sort(vc.begin(),vc.end(),cmp);
// for(node t : vc){
// cout<<t.l<<" "<<t.r<<"\n";
// }
queue<int>qu;
int cnt = 1, cc = 1;
qu.push(vc[0].r);
for(int i = 1; i < vc.size(); i++){
if(vc[i].l > qu.front()){
while(vc[i].l > qu.front()&&qu.size()>1){
qu.pop();
cc--;
}
cnt++;
cc++;
qu.push(vc[i].r);
//cout<<vc[i].l<<" "<<vc[i].r<<" "<<cc<<22<<'\n';
}else{
while(cc>=k&&qu.size()>1&&vc[i].l>qu.front()){
qu.pop();
cc--;
}
if(cc<k){
cnt++;
qu.push(vc[i].r);
cc++;
//cout<<vc[i].l<<" "<<vc[i].r<<" "<<cc<<'\n';
}
}
}
cout<<cnt<<"\n";
return 0;
}J、集合论
- 分四类讨论结论题，二进制函数没开ll卡了很久。
#include<bits/stdc++.h>
#define rep(i,x,y) for(auto i=(x);i<=(y);++i)
#define dep(i,x,y) for(auto i=(x);i>=(y);--i)
#define SCC(x) scanf("%d",&x)
#define ISS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
typedef long long ll;
using namespace std;
const int N=1e6+5;
const int D=1e7+5;
const ll MOD=998244353;
const double Pi = acos(-1.0);
//inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
//inline void write(int x){if (x < 0) x = ~x + 1, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + '0');}
//ll pow1(ll a,ll b){ll r=1;while(b){if(b&1)r=r*a;a=a*a;b>>=1;}return r;}
//ll pow2(ll a,ll b,ll mod){ll r=1;while(b){if(b&1)r=r*a%mod;a=a*a%mod;b>>=1;}return r%mod;}
//ll gcd(ll a,ll b){ll m=a%b;while(m){a=b;b=m;m=a%b;}return b;}
//void exgcd(ll a, ll b, ll &d, ll &x, ll &y){if(!b) { d = a; x = 1; y = 0;}else { exgcd(b, a % b, d, y, x); y -= x * (a / b); }}
//ll inv(ll a){ll d, x, y;exgcd(a, MOD, d, x, y);return d == 1 ? (x + MOD) % MOD : -1;}struct node{ll mat[105][105];};node mul(node x,node y,int n,ll mod){node tmp;for(int i=0;i<n;i++)for(int j=0;j<n;j++){tmp.mat[i][j]=0;for(int k=0;k<n;k++)tmp.mat[i][j]+=(x.mat[i][k]*y.mat[k][j])%mod;tmp.mat[i][j]%=mod;}return tmp;}
// node matpow(node x,node y,ll num,int n,ll mod){while(num){if(num&1){y=mul(x,y,n,mod);}x=mul(x,x,n,mod);num=num>>1;}return y;}
//for(int i=2;i<D;i++){if(!np[i]) p[++cnt]=i;for(int j=1;j<=cnt&&p[j]*i<D;j++){np[i*p[j]]=1;if(i%p[j]==0) break;}}
//int find(int x){return a[x]==x?x:a[x]=find(a[x]);}void merge(int x,int y){a[find(y)]=a[find(x)];}
//string s1="0",s2="{0}";
set<ll>t1,t2,t3,t4;
ll highbit(ll x) { return (ll)1 << (63 - __builtin_clzll(x)); }
int main(){
// cout<<s1<<" "<<s1.size()<<'\n'<<s2<<" "<<s2.size()<<'\n';
// rep(i,1,10){
// s1=s1+","+s2;
// s2="{"+s1+"}";
// cout<<s2<<" "<<s2.size()<<'\n';
// }
// int x;
// cin>>x;
ll x,t;
x=4;rep(i,1,60) t1.insert(x-1),x*=2;
x=1;rep(i,1,62) t2.insert(x),x*=2;t2.erase(2);
t3.insert(2);
x=8;rep(i,1,60) t4.insert(x-2),x*=2;
cin>>t;
while(t--){
cin>>x;
while(1){
if(t1.count(x)) {cout<<','<<'\n';break;}
if(t2.count(x)) {cout<<'{'<<'\n';break;}
if(t3.count(x)) {cout<<'0'<<'\n';break;}
if(t4.count(x)) {cout<<'}'<<'\n';break;}
x=x-highbit(x)+1;
}
}
}