2018 icpc 沈阳

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;
}
View Code

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;
}
View Code

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;
}
View Code

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;
}
View Code

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;
}
View Code

 

posted @ 2019-09-15 21:24  ACforever  阅读(1013)  评论(0编辑  收藏  举报