蓝桥杯第十届省赛A组D~E题（快速乘）

测试用例如下：

01010101001011001001010110010110100100001000101010

00001000100000101010010000100000001001100110100101

01111011010010001000001101001011100011000000010000

01000000001010100011010000101000001010101011001011

00011111000000101000010010100010100000101100000000

11001000110101000010101100011010011010101011110111

00011011010101001001001010000001000101001110000000

10100000101000100110101010111110011000010000111010

00111000001010100001100010000001000101001100001001

11000110100001110010001001010101010101010001101000

00010000100100000101001010101110100010101010000101

11100100101001001000010000010101010100100100010100

00000010000000101011001111010001100000101010100011

10101010011100001000011000010110011110110100001000

10101010100001101010100101000010100000111011101001

10000000101100010000101100101101001011100000000100

10101001000000010100100001000100000100011110101001

00101001010101101001010100011010101101110000110101

11001010000100001100000010100101000001000111000010

00001000110000110101101000000100101001001000011101

10100101000101000000001110110010110101101010100001

00101000010000110101010000100010001001000100010101

10100001000110010001000010101001010101011111010010

00000100101000000110010100101001000001000000000010

11010000001001110111001001000011101001011011101000

00000110100010001000100000001000011101000000110011

10101000101000100010001111100010101001010000001000

10000010100101001010110000000100101010001011101000

00111100001000010000000110111000000001000000001011

10000001100111010111010001000110111010101101111000

 

 本题需要用BFS来求解。（因为要求最短路。如果单纯的求最小的字典序的话就需要用DFS了）。

用BFS求解的话可以得到到达终点的最短路。但如果要求最小的字典序的话，我们只需要按照DLRU这个顺序依次入列该点周边的点即可。

AC代码如下：

#include<bits/stdc++.h>
#define MAXN 100005
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
char mg[35][55];
int ans[10000],vis[35][55],cnt;
int to[4][2]={1,0, 0,-1, 0,1, -1,0};//D,L,R,U
char d[4]={'D','L','R','U'};
struct node{
    int x,y;
    string ans;//到达该点所需要的最短路径的最小字典序的方位合集 
    node(int _x,int _y,string _ans){
        x=_x,y=_y,ans=_ans;
    }
};
int main(){
    for(int i=1;i<=30;++i)
        scanf("%s",mg[i]+1);
    queue<node>q;
    q.push(node(1,1,""));
    while(!q.empty()){
        node u=q.front();
        q.pop();
        vis[u.x][u.y]=1;
        if(u.x==30&&u.y==50){
            cout<<u.ans;
            return 0;
        }
        for(int i=0;i<4;i++){
            int tempx=u.x+to[i][0],tempy=u.y+to[i][1];
            if(tempx==0||tempx==31||tempy==0||tempy==51||vis[tempx][tempy]||mg[tempx][tempy]=='1')
                continue;
            else
                q.push(node(tempx,tempy,u.ans+d[i]));
            
        }
    }
    return 0;
}

答案如下：

DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR

 

先暴力求解出q和p。再根据exgcd求解出e。最后根据快速幂和快速乘求解出答案即可。（一定要用快速乘不然会溢出）

AC代码如下：

#include<bits/stdc++.h>
#define MAXN 100005
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
char mg[35][55];
ll n=1001733993063167141,d=212353,c=20190324,p,q,x,y;
void exgcd(ll a,ll b){
    if(!b){
        x=1,y=0;
        return ;
    }
    exgcd(b,a%b);
    ll t=x;
    x=y;
    y=t-a/b*y;
    return ;
}
ll mul(ll a,ll b){//快速乘，防止溢出 
    ll ans=0;
    while(b){
        if(b&1)
            ans=(ans+a)%n;
        a=(a*2)%n;
        b>>=1; 
    } 
    return ans;
}
ll quick_mod(ll c,ll e){//快速幂 
    ll ans=1;
    while(e){
        if(e&1)
            ans=mul(ans,c)%n;
        c=mul(c,c)%n;
        e>>=1;
    }
    return ans;
}
int main(){
    for(int i=2;i<sqrt(n);i++)
        if(n%i==0){
            p=i,q=n/i;
            break;
        }
    printf("p:%lld,q:%lld\n",p,q);
    ll m =(p-1)*(q-1);
    exgcd(d,m);
    x=(x%m+m)%m;
    printf("x:%lld\n",x);
    ll ans=quick_mod(c,x);
    printf("ans:%lld\n",ans);
    return 0;
}