JNday2-pm

 

今天下午的题目感觉比较简单（前两题），第三题很难

T1在北京时做过，然后就几分钟了出来

T2貌似也是一个暴力题，以为不打素数表会超时（其实不然），然而打表超代码内存，连交都交不上去，还真不知道代码内存是多少

T3差不多做了将近1.5h，然而题目真的很难读懂的，还有半个小时才知道屏幕上有样例解释（看不清楚）GG，放弃

T1

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>

using namespace std;
const int N = 1e5 + 10;

string s; 
int answer = 0;
int stack[N], l, r;

int main()
{
    freopen("shower.in", "r", stdin);
    freopen("shower.out", "w", stdout);
    cin >> s;
    int len = s.length();
    l = 1, r = 0;
    for(int i = 0; i < len; i ++)
    {
        if(s[i] == '(')
            stack[++ r] = 1;
        else
        {
            if(l > r)
            {
                answer ++;
                stack[++ r] = 1;
            }
            else
                r --;
        }
    }
    answer += (r >> 1);
    printf("%d", answer);
    return 0;
}
/*
)()()()()(()()()((
*/

T2

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 1000005;

#define LL long long

int s[N];

inline int read()
{
    int x = 0; char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x;
}

int T, n, k, js;
LL f[N];
bool vis[1000005], flag;

inline bool check(int x)
{
    LL ret = f[x + k - 1] - f[x - 1];
    if(ret <= n) return 1;
    else return 0;
}

inline void make_s()
{
    int ove = 1000005;
    for(int i = 2; i <= sqrt(ove) + 1; i ++) 
        if(vis[i] == 0) 
            for(int j = i * 2; j <= ove; j += i) 
                vis[j] = 1;
    for(int i = 2; i <= ove; i ++) if(!vis[i]) s[++ js] = i;
}

int main()
{
    //freopen("diary.in", "r", stdin);
    //freopen("bbbbb.out", "w", stdout);
    make_s();
    for(int i = 1; i <= js; i ++) 
    {
        f[i] = f[i - 1] + s[i];
    }
    T = read();
    while(T --)
    {
        n = read();
        k = read();
        int L = 1, R = js, mid, Ans;
        while(L <= R)
        {
            mid = (L + R) >> 1;
            if(check(mid)) L = mid + 1, Ans = mid;
            else R = mid - 1;
        }
        if(R == 0)
        {
            printf("-1\n");
            continue ;
        }
        int answer = f[Ans + k - 1] - f[Ans - 1];
        if(answer > n) printf("-1\n");
        else printf("%lld\n", answer);
    }    
    return 0;
}
/*
3
20 2
20 3
20 4
1
659309 1
659299
*/

T3

#include <algorithm>
#include <cstdio>
#include <map>
#define mp std::make_pair
const int N = 65;
const int mod = 1000000007;
typedef long long LL;
typedef std::map<std::pair<LL, LL>, LL> Mapd;
typedef std::map<LL, LL> Mapf;
Mapd mapd[N];
Mapf mapf[N];
int a[N], b[N];
LL F[N], c[N], d[N], n[N], l[N];
LL dd(int i, LL x, LL y)
{
    if (x > y) std::swap(x, y);
    if (x == y) return 0;
    Mapd::iterator it = mapd[i].find(mp(x, y));
    if (it != mapd[i].end()) return it->second;
    LL na = n[a[i]];
    if (x >= na)
        return mapd[i][mp(x, y)] = dd(b[i], x - na, y - na);
    else if (y < na)
        return mapd[i][mp(x, y)] = dd(a[i], x, y);
    else
        return mapd[i][mp(x, y)] = (dd(a[i], x, c[i]) + dd(b[i], d[i], y - na) + l[i]) % mod;
}
LL f(int i, LL j)
{
    if (!i) return 0;
    Mapf::iterator it = mapf[i].find(j);
    if (it != mapf[i].end()) return it->second;
    LL na = n[a[i]];
    if (j < na)
        return mapf[i][j] =
                   (f(a[i], j) + n[b[i]] % mod * dd(i, j, d[i] + na) % mod + f(b[i], d[i])) % mod;
    else
        return mapf[i][j] =
                   (f(b[i], j - na) + n[a[i]] % mod * dd(i, j, c[i]) % mod + f(a[i], c[i])) % mod;
}
int main()
{
    freopen("cloth.in", "r", stdin);
    freopen("cloth.out", "w", stdout);
    int m;
    scanf("%d", &m);
    F[0] = 0;
    n[0] = 1;
    for (int i = 1; i <= m; ++i)
    {
        scanf("%d%d%I64d%I64d%I64d", &a[i], &b[i], &c[i], &d[i], &l[i]);
        n[i] = n[a[i]] + n[b[i]];
        LL na = n[a[i]] % mod, nb = n[b[i]] % mod;
        l[i] %= mod;
        printf("%I64d\n", F[i] =
                   (F[a[i]] + F[b[i]] +
                    nb * f(a[i], c[i]) % mod +
                    na * f(b[i], d[i]) % mod +
                    na * nb % mod * l[i] % mod) % mod);
    }
    return 0;
}