https://codeforces.com/gym/101955
J
签到
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#define ll long long
#define fo(i,l,r) for(int i = l;i <= r;i++)
#define fd(i,l,r) for(int i = r;i >= l;i--)
using namespace std;
const int maxn = 5e5+5;
ll read() {
ll 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 * 10 + (ch - '0');
ch = getchar();
};
return x*f;
}
char s[105];
int main(){
int T;
cin>>T;
int n;
ll ans,a,b;
int Case = 0;
while(T--){
n=read();
ans = 0;
fo(tt,1,n) {
cin.getline(s + 1, 100);
b = a = 0;
if (s[1] == 'b') a = 1;
if (s[1] == 'c') a = 1;
if (s[1] == 'i') a = 4;
if (s[1] == 'l' && s[6] == 'l')a = 8;
if (s[1] == '_')a = 16;
if (s[1] == 'f')a = 4;
if (s[1] == 'd')a = 8;
if (!a)a = 16;
int len = strlen(s + 1);
int bs = 1;
if (s[len - 1] == ']') {
for (int i = len - 2; i >= 1; i--) {
if (s[i] == '[')break;
b = b + (s[i] - '0') * bs;
bs *= 10;
}
} else {
b = 1;
}
ans += a * b;
}
if(ans%1024) ans = ans/1024 + 1;
else ans = ans/1024;
cout<<"Case #"<<++Case<<": "<<ans<<endl;
}
return 0;
}
C
先乘一个k!，并假定前k个数是有序的
最长上升序列长度是n-1，就要把一个数字排除在外，就要把一个数字插入到其他位置
分成三种情况：前k个数里的某个数插入到后n-k个里，后n-k个数插入到前k个数里，后n-k个数插入到后n-k个数里。
分别统计，要注意相邻两个数交换的情况，只能被统计一次。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#define ll long long
#define fo(i,l,r) for(int i = l;i <= r;i++)
#define fd(i,l,r) for(int i = r;i >= l;i--)
using namespace std;
const int maxn = 505;
ll read() {
ll 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 * 10 + (ch - '0');
ch = getchar();
};
return x*f;
}
int main(){
int T;
T=read();
ll n,k,mod;
ll ans,a,b;
int Case = 0;
while(T--){
n=read();
k=read();
mod=read();
k = min(n,k);
a=1;
for(ll i = 1;i <= k;i++){
a*=i;
a%=mod;
}
b=(n-1)*(n-k)+1;
b%=mod;
ans = (a*b)%mod;
cout<<"Case #"<<++Case<<": "<<ans<<endl;
}
return 0;
}
G
圆上的整数点数量是相当有限的，对于这个题来说，询问的半径是固定不变的，可以预处理圆上的整点，由于点的数量很小，接下来就暴力修改查询即可。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include <set>
#include <map>
#include <queue>
#include <cmath>
#define ll long long
#define fo(i,l,r) for(int i = l;i <= r;i++)
#define fd(i,l,r) for(int i = r;i >= l;i--)
using namespace std;
const int maxn = 6050,maxk=1e7+5;
ll read() {
ll 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 * 10 + (ch - '0');
ch = getchar();
};
return x*f;
}
const int fx[4] = {-1,1,-1,1};
const int fy[4] = {-1,-1,1,1};
vector<int> st[maxk],st2[maxk];
int n,m,flag;
ll weight[maxn+10][maxn+10];
int flags[maxn+10][maxn+10];
bool judge(int y,int x){
return y>=0&&y<=maxn&&x>=0&&x<=maxn;
}
void add(int y,int x,ll w){
weight[y][x] = w;
flags[y][x] = flag;
}
void del(int y,int x){
weight[y][x] = 0;
flags[y][x]--;
}
void inc(int y,int x,ll k,ll w){
int sz = st[k].size();
int dy,dx,nowy,nowx;
fo(i,0,sz-1){
dx = st[k][i];
dy = st2[k][i];
fo(j,0,3){
if(dx==0&&(j==1||j==3))continue;
if(dy==0&&(j==2||j==3))continue;
nowy = y + dy*fy[j];
nowx = x + dx*fx[j];
if(!judge(nowy,nowx))continue;
if(flags[nowy][nowx]==flag) weight[nowy][nowx] += w;
}
}
}
ll query(int y,int x,ll k){
int sz = st[k].size();
int dy,dx,nowy,nowx;
ll ans = 0;
fo(i,0,sz-1){
dx = st[k][i];
dy = st2[k][i];
fo(j,0,3){
if(dx==0&&(j==1||j==3))continue;
if(dy==0&&(j==2||j==3))continue;
nowy = y + dy*fy[j];
nowx = x + dx*fx[j];
if(judge(nowy,nowx)){
if(flags[nowy][nowx]==flag)ans += weight[nowy][nowx];
}
}
}
return ans;
}
int main(){
ios::sync_with_stdio(false);
int T,Case=0;
T=read();
ll lastans = 0;
fo(i, 0, 6000) {
fo(j, 0, 6000) {
if (i * i + j * j > 1e7) break;
st[i * i + j * j].push_back(i);
st2[i * i + j * j].push_back(j);
}
}
while(T--) {
flag++;
lastans=0;
int y, x, cmd;
ll k, w;
cout<<"Case #"<<++Case<<":"<<endl;
n = read();
m = read();
fo(i, 1, n) {
y = read();
x = read();
w = read();
add(y, x, w);
}
fo(i, 1, m) {
cmd = read();
y = read();
x = read();
y = (y + lastans) % 6000ll + 1ll;
x = (x + lastans) % 6000ll + 1ll;
if (cmd == 1) {
w = read();
add(y, x, w);
} else if (cmd == 2) {
del(y, x);
} else if (cmd == 3) {
k = read();
w = read();
inc(y, x, k, w);
} else {
k = read();
cout << (lastans = query(y, x, k)) << endl;
}
}
}
return 0;
}
K
首先考虑一般意义上的约瑟夫环问题，把k-1个人删去后，将k...n,0..k-2重新编号，就变成一个规模为n-1的一模一样的问题了。
考虑这个题，这次求的不是最后一个，而是第m个人把k-1删去之后，就变成了从n-1个人里找第m-1个人了。
直接递推需要一个o(m)的循环，但是题目范围说m和k总有一个很小，当k很小m很大时，我们可以让它直接顶到头而不是一项一项递推，加速递推过程。
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <set>
#include <cmath>
#include <queue>
#include <map>
#define ll long long
#define ld long double
#define lson rt << 1, l, m
#define pi acos(-1)
#define rson rt << 1 | 1, m + 1, r
#define fo(i, l, r) for (long long i = l; i <= r; i++)
#define fd(i, l, r) for (long long i = r; i >= l; i--)
#define mem(x) memset(x, 0, sizeof(x))
#define eps 3e-11
using namespace std;
const ll maxn = 40000050;
const ll mod = 998244353;
ll read()
{
ll 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 * 10 + (ch - '0');
ch = getchar();
};
return x * f;
}
ll n, m, k;
int main()
{
int T;
T = read();
int tt = 0;
while (T--)
{
tt++;
n = read();
m = read();
k = read();
ll ans = (k - 1) % (n - m + 1);
if (k == 1)
{
ans = m - 1;
}
else
{
fo(i, n - m + 2, n)
{
ans = (ans + k) % i;
ll js = i - ans - 1;
js /= k;
js--;
if (n - i - 1 < js)
js = n - i - 1;
if (js > 0)
{
i += js;
ans = ans + k * js;
}
}
}
printf("Case #%d: %I64d\n", tt, ans + 1);
}
return 0;
}
L
计算几何。
首先将外部圆与中心圆的交点求出，先看能不能取到直径，可以取到的充要条件是，存在一个交点，它的对称点没有被外部圆切去。
若不能取到直径，所有交点两两连线，去距离的最大值。
本地精度存在一些问题，long double会导致WA，只有double可以过，具体原因待探究。
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <set>
#include <cmath>
#include <queue>
#include <map>
#define ll long long
#define ld double
#define lson rt << 1, l, m
#define pi acos(-1)
#define rson rt << 1 | 1, m + 1, r
#define fo(i, l, r) for (long long i = l; i <= r; i++)
#define fd(i, l, r) for (long long i = r; i >= l; i--)
#define mem(x) memset(x, 0, sizeof(x))
#define eps 1e-10
using namespace std;
const ll maxn = 1050;
const ll mod = 998244353;
ll read()
{
ll 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 * 10 + (ch - '0');
ch = getchar();
};
return x * f;
}
struct dat
{
ld v;
bool isr;
int p;
friend bool operator<(dat a, dat b)
{
return a.v < b.v;
}
};
int cnt;
ll n;
ld R;
ld px[maxn], py[maxn], pr[maxn], ans;
ld fpx[maxn], fpy[maxn];
dat ang[maxn * 2];
dat fang[maxn * 2];
ld getang(ld a, ld b, ld c)
{
return acos((a * a + b * b - c * c) / (a * b + a * b));
}
ld tryang(ld agl)
{
if (agl < 0)
agl = -agl;
return R * sin(agl / 2.0) * 2.0;
}
int main()
{
int T;
T = read();
int tt = 0;
while (T--)
{
tt++;
cnt = 0;
n = read();
R = read();
ld dis, baseAngle;
fo(i, 1, n)
{
px[i] = read();
py[i] = read();
pr[i] = read();
dis = sqrt(px[i] * px[i] + py[i] * py[i]);
baseAngle = atan2(py[i], px[i]);
if (pr[i] + R < dis)
{
continue;
}
if (fabs(R - pr[i]) > dis)
{
continue;
}
ang[cnt + 1].v = baseAngle - getang(dis, R, pr[i]);
ang[cnt + 2].v = baseAngle + getang(dis, R, pr[i]);
if (ang[cnt + 1].v < 0)
ang[cnt + 1].v += pi + pi;
if (ang[cnt + 1].v >= pi + pi)
ang[cnt + 1].v -= pi + pi;
if (ang[cnt + 2].v < 0)
ang[cnt + 2].v += pi + pi;
if (ang[cnt + 2].v >= pi + pi)
ang[cnt + 2].v -= pi + pi;
cnt += 2;
}
if (cnt == 0)
ans = R + R;
fo(i, 1, cnt)
{
fpx[i] = -R * cos(ang[i].v);
fpy[i] = -R * sin(ang[i].v);
}
ans = 0.0;
fo(i, 1, cnt)
{
fo(j, i + 1, cnt)
{
ans = max(ans, tryang(ang[i].v - ang[j].v));
}
}
ld tx, ty;
fo(i, 1, cnt)
{
bool ok = false;
fo(j, 1, n)
{
if ((fpx[i] - px[j]) * (fpx[i] - px[j]) + (fpy[i] - py[j]) * (fpy[i] - py[j]) < pr[j] * pr[j])
{
ok = true;
break;
}
}
if (!ok)
{
ans = R + R;
break;
}
}
printf("Case #%d: %.15lf\n", tt, ans);
}
return 0;
}