2019dx#7

Solved	Pro.ID	Title	Ratio(Accepted / Submitted)
	1001	A + B = C	10.48%(301/2872)
	1002	Bracket Sequences on Tree	11.27%(16/142)
	1003	Cuber Occurrence	6.67%(1/15)
	1004	Data Structure Problem	23.08%(3/13)
	1005	Equation	0.00%(0/63)
	1006	Final Exam　　　　　　　　　　推公式，田忌赛马	5.06%(297/5872)
	1007	Getting Your Money Back	12.42%(20/161)
	1008	Halt Hater	14.77%(61/413)
	1009	Intersection of Prisms	0.00%(0/2)
	1010	Just Repeat　　　　　　　　　　博弈，贪心	15.04%(128/851)
	1011	Kejin Player　　　　　　　　　　期望DP	21.20%(544/2566)

1001 A + B = C

题意：

给定a,b,c($a, b, c \le 10 ^{100000}$)，求一组x, y, z满足$a \times 10^x + b \times 10^y = c \times 10^z$ 。

思路：

先把每个数末尾的0去掉，然后可以发现满足如下等式之一$$ a + b = c \times 10 ^k $$ $$ a \times 10 ^k + b = c $$ $$ a + b \times 10 ^ k = c$$就行了。

由于是大数，可以利用哈希，计算等式中的k，可以移项，乘逆元，预处理mod意义下指数。

然后我用到双哈希保险

// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #pragma GCC optimize(4)
#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
#include <unordered_map>
// #include<bits/extc++.h>
// using namespace __gnu_pbds;
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
/**********showtime************/
            const int maxn = 1e5+;
            const int N = 1e5+;
            char a[maxn],b[maxn],c[maxn];
            char d[maxn];
            int mod1 = , mod2 = 1e9+;
            int ten1[N], ten2[N];
            unordered_map<int,int>mp1,mp2;
            void init(){
                ten1[] = ten2[] = ;
                mp1[] = mp2[] = ;
                for(int i=; i<N; i++)
                {
                    ten1[i] = 1ll*ten1[i-] *  % mod1;
                    ten2[i] = 1ll*ten2[i-] *  % mod2;

                    mp1[ten1[i]] = i;
                    mp2[ten2[i]] = i;
                }
            }
            ll ksm(ll a, ll b, ll mod){
                ll res = ;
                while(b > ) {
                    if(b & ) res = res * a % mod;
                    a = a * a % mod;
                    b = b >> ;
                }
                return res;
            }
int main(){
            init();
//            debug("ok");
            int T;  scanf("%d", &T);

            while(T--) {
                scanf("%s%s%s", a, b, c);
                int alen = strlen(a), blen = strlen(b), clen = strlen(c);
                int cnta = , cntb = , cntc = ;
                while(a[alen-] == '') alen--, cnta ++;
                while(b[blen-] == '') blen--, cntb ++;
                while(c[clen-] == '') clen--, cntc ++;
                int kk = max(cnta, max(cntb, cntc));
                a[alen] = '\0';
                b[blen] = '\0';
                c[clen] = '\0';
                ll A1 = , B1 = , C1 = ;
                ll A2 = , B2 = , C2 = ;

                for(int i=; i<alen; i++){
                    A1 = (A1 *  + (a[i] - '') ) % mod1;
                    A2 = (A2 *  + (a[i] - '') ) % mod2;
                }

                for(int i=; i<blen; i++){
                    B1 = (B1 *  + (b[i] - '') ) % mod1;
                    B2 = (B2 *  + (b[i] - '') ) % mod2;
                }

                for(int i=; i<clen; i++){
                    C1 = (C1 *  + (c[i] - '') ) % mod1;
                    C2 = (C2 *  + (c[i] - '') ) % mod2;
                }

                int k1 = (A1 + B1) % mod1 * ksm(C1, mod1-, mod1) % mod1;
                if(mp1.count(k1))k1 = mp1[k1];
                else k1 = -;
                int k2 = (A2 + B2) % mod2 * ksm(C2, mod2-, mod2) % mod2;
                if(mp2.count(k2))k2 = mp2[k2];
                else k2 = -;
                if(k1 == k2 && k1 >= ) {
                    printf("%d %d %d\n", kk - cnta, kk - cntb, kk + k1 - cntc);
                    continue;
                }

                k1 = (C1 - B1) % mod1;
                if(k1 < ) k1 += mod1;
                k1 = k1 * ksm(A1, mod1-, mod1) % mod1;
                if(mp1.count(k1))k1 = mp1[k1];
                else k1 = -;

                k2 = (C2 - B2) % mod2;
                if(k2 < ) k2 += mod2;
                k2 = k2 * ksm(A2, mod2-, mod2) % mod2;
                if(mp2.count(k2))k2 = mp2[k2];
                else k2 = -;
                if(k1 == k2 && k1 >= ) {
                    printf("%d %d %d\n", kk + k1 - cnta, kk - cntb, kk - cntc);
                    continue;
                }

                k1 = (C1 - A1) % mod1;
                if(k1 < ) k1 += mod1;
                k1 = k1 * ksm(B1, mod1-, mod1) % mod1;
                if(mp1.count(k1))k1 = mp1[k1];
                else k1 = -;

                k2 = (C2 - A2) % mod2;
                if(k2 < ) k2 += mod2;
                k2 = k2 * ksm(B2, mod2-, mod2) % mod2;
                if(mp2.count(k2))k2 = mp2[k2];
                else k2 = -;
                if(k1 == k2 && k1 >= ) {
                    printf("%d %d %d\n", kk  - cnta, kk  + k1 - cntb, kk - cntc);
                    continue;
                }
                puts("-1");
            }
            return ;
}

1006 Final Exam

思路：

假设每个题目所用的时间为$a_1, a_2, ... , a_n(a_i <= a_{i+1})$

按老师的想法，为了不让学生过掉n - k 个题目，肯定是把$a_1, a_2,...a_{n-k}$ 对应题目的分值设为$a_1, a_2,...a_{n-k}$.

然后$$m - a_1 - a_2 - ... - a_{n-k} < a_{n-k+1}$$

我们给左边+1,再移项，变成

$$m + 1 \le + a_1 + a_2 + ... + a_{n-k} + a_{n-k+1}$$

可以发现$a_{n-k+1}$最大会被老师卡成$\left\lceil  \frac{m + 1}{n - k + 1} \right\rceil$

之后的$a_{n-k+2},,,a_{n}$可以等于$a_{n-k+1}$就行了。

// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #pragma GCC optimize(4)
#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
// #include<bits/extc++.h>
// using namespace __gnu_pbds;
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
/**********showtime************/

int main(){
            int T;  scanf("%d", &T);
            while(T--) {
                ll n, m , k;
                scanf("%lld%lld%lld", &n, &m, &k);
                ll tmp = (m + ) / (n - k + );
                if((m+) % (n - k + )) tmp++;
                ll ans = (k - ) * tmp + m+ ;
                printf("%lld\n", ans);
            }

            return ;
}

1008 Halt Hater

题意：

一开始你在(0, 0)点，面向Y轴正方向。向左走费用为a，向前走费用为b，向右走费用为0。有T（$\le 100000$） 组数据，每组给定x，y，a，b，问你到（x，y）的最小费用。

思路：

规律题，首先发现，你到（x-1, y+1）(x, y+1), (x-1, y), (x, y) 其中之一就行了。

然后发现，如果把每个格子看成一个点，那么，相邻格子的费用为a。斜对着的两个格子的费用为min（a, 2 * b）。

一下跨两步的费用为min(2*a, 2*b)。

所以我们首先斜着走，然后两步两步走，然后再走一步。

// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #pragma GCC optimize(4)
#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
#include <unordered_map>
// #include<bits/extc++.h>
// using namespace __gnu_pbds;
using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}
/**********showtime************/

            ll a, b, x, y;
            ll check(ll x, ll y) {

                x = abs(x), y = abs(y);

                ll ans = ;
                ll m = min(x, y);

                ans += m * min(a, 2ll * b);
                ll yu = x + y - m - m;
                ans += (yu / ) * min(2ll * a , 2ll * b);

                if(yu %  == ) ans += b;
                return ans;
            }
int main(){
            int T;  scanf("%d", &T);
            while(T--) {
                scanf("%lld%lld%lld%lld", &a, &b, &x, &y);
                ll ans = check(x, y);
                ans = min(ans, check(x-, y));
                ans = min(ans, check(x, y+));
                ans = min(ans, check(x-, y+));
                printf("%lld\n", ans);
            }
            return ;
}