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 
#include  
#include  
#include   
#include   
#include   
#include    
#include    
#include    
#include    <string>
#include    
#include     
#include     
#include     
#include      
#include       
#include       <set>
#include   
#include 
// #include
// 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<<"-----------------------"<#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 pll;

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;

template
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
/**********showtime************/
            const int maxn = 1e5+9;
            const int N = 1e5+9;
            char a[maxn],b[maxn],c[maxn];
            char d[maxn];
            int mod1 = 998244353, mod2 = 1e9+7;
            int ten1[N], ten2[N];
            unordered_map<int,int>mp1,mp2;
            void init(){
                ten1[0] = ten2[0] = 1;
                mp1[1] = mp2[1] = 0;
                for(int i=1; i)
                {
                    ten1[i] = 1ll*ten1[i-1] * 10 % mod1;
                    ten2[i] = 1ll*ten2[i-1] * 10 % mod2;

                    mp1[ten1[i]] = i;
                    mp2[ten2[i]] = i;
                }
            }
            ll ksm(ll a, ll b, ll mod){
                ll res = 1;
                while(b > 0) {
                    if(b & 1) res = res * a % mod;
                    a = a * a % mod;
                    b = b >> 1;
                }
                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 = 0, cntb = 0, cntc = 0;
                while(a[alen-1] == '0') alen--, cnta ++;
                while(b[blen-1] == '0') blen--, cntb ++;
                while(c[clen-1] == '0') clen--, cntc ++;
                int kk = max(cnta, max(cntb, cntc));
                a[alen] = '\0';
                b[blen] = '\0';
                c[clen] = '\0';
                ll A1 = 0, B1 = 0, C1 = 0;
                ll A2 = 0, B2 = 0, C2 = 0;

                for(int i=0; i){
                    A1 = (A1 * 10 + (a[i] - '0') ) % mod1;
                    A2 = (A2 * 10 + (a[i] - '0') ) % mod2;
                }

                for(int i=0; i){
                    B1 = (B1 * 10 + (b[i] - '0') ) % mod1;
                    B2 = (B2 * 10 + (b[i] - '0') ) % mod2;
                }

                for(int i=0; i){
                    C1 = (C1 * 10 + (c[i] - '0') ) % mod1;
                    C2 = (C2 * 10 + (c[i] - '0') ) % mod2;
                }

                int k1 = (A1 + B1) % mod1 * ksm(C1, mod1-2, mod1) % mod1;
                if(mp1.count(k1))k1 = mp1[k1];
                else k1 = -1;
                int k2 = (A2 + B2) % mod2 * ksm(C2, mod2-2, mod2) % mod2;
                if(mp2.count(k2))k2 = mp2[k2];
                else k2 = -1;
                if(k1 == k2 && k1 >= 0) {
                    printf("%d %d %d\n", kk - cnta, kk - cntb, kk + k1 - cntc);
                    continue;
                }

                k1 = (C1 - B1) % mod1;
                if(k1 < 0) k1 += mod1;
                k1 = k1 * ksm(A1, mod1-2, mod1) % mod1;
                if(mp1.count(k1))k1 = mp1[k1];
                else k1 = -1;

                k2 = (C2 - B2) % mod2;
                if(k2 < 0) k2 += mod2;
                k2 = k2 * ksm(A2, mod2-2, mod2) % mod2;
                if(mp2.count(k2))k2 = mp2[k2];
                else k2 = -1;
                if(k1 == k2 && k1 >= 0) {
                    printf("%d %d %d\n", kk + k1 - cnta, kk - cntb, kk - cntc);
                    continue;
                }

                k1 = (C1 - A1) % mod1;
                if(k1 < 0) k1 += mod1;
                k1 = k1 * ksm(B1, mod1-2, mod1) % mod1;
                if(mp1.count(k1))k1 = mp1[k1];
                else k1 = -1;

                k2 = (C2 - A2) % mod2;
                if(k2 < 0) k2 += mod2;
                k2 = k2 * ksm(B2, mod2-2, mod2) % mod2;
                if(mp2.count(k2))k2 = mp2[k2];
                else k2 = -1;
                if(k1 == k2 && k1 >= 0) {
                    printf("%d %d %d\n", kk  - cnta, kk  + k1 - cntb, kk - cntc);
                    continue;
                }
                puts("-1");
            }
            return 0;
}
View Code

 

 

 

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 
#include  
#include  
#include   
#include   
#include   
#include    
#include    
#include    
#include    <string>
#include    
#include     
#include     
#include     
#include      
#include       
#include       <set>
#include   
// #include
// 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<<"-----------------------"<#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 pll;

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;

template
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',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 + 1) / (n - k + 1);
                if((m+1) % (n - k + 1)) tmp++;
                ll ans = (k - 1) * tmp + m+ 1;
                printf("%lld\n", ans);
            }

            return 0;
}
View Code

 

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 
#include  
#include  
#include   
#include   
#include   
#include    
#include    
#include    
#include    <string>
#include    
#include     
#include     
#include     
#include      
#include       
#include       <set>
#include   
#include 
// #include
// 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<<"-----------------------"<#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 pll;

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;

template
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',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 = 0;
                ll m = min(x, y);

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

                if(yu % 2 == 1) 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-1, y));
                ans = min(ans, check(x, y+1));
                ans = min(ans, check(x-1, y+1));
                printf("%lld\n", ans);
            }
            return 0;
}
View Code

 

转载于:https://www.cnblogs.com/ckxkexing/p/11341725.html

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