Educational Codeforces Round 63 (Rated for Div. 2) -E

E. Guess the Root
time limit per test:2 seconds
memory limit per test:256 megabytes
inputstandard input
outputstandard output
题目链接：http://codeforces.com/contest/1155/problem/E
Jury picked a polynomial f(x)=a0+a1⋅x+a2⋅x2+⋯+ak⋅xk. k≤10 and all ai are integer numbers and 0≤ai<106+3. It’s guaranteed that there is at least one i such that ai>0.

Now jury wants you to find such an integer x0 that f(x0)≡0mod(106+3) or report that there is not such x0.

You can ask no more than 50 queries: you ask value xq and jury tells you value f(xq)mod(106+3).

Note that printing the answer doesn’t count as a query.

Interaction
To ask a question, print “? xq” (0≤xq<106+3). The judge will respond with a single integer f(xq)mod(106+3). If you ever get a result of −1 (because you printed an invalid query), exit immediately to avoid getting other verdicts.

After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:

fflush(stdout) or cout.flush() in C++;
System.out.flush() in Java;
flush(output) in Pascal;
stdout.flush() in Python;
see documentation for other languages.
When you are ready to answer, print “! x0” where x0 is the answer or −1 if there is no such x0.

You can ask at most 50 questions per test case.

Hack Format

To hack, use the following format.

The only line should contain 11 integers a0,a1,…,a10 (0≤ai<106+3, max0≤i≤10ai>0) — corresponding coefficients of the polynomial.

Examples
inputCopy

1000002

0
outputCopy
? 0

? 1

! 1
inputCopy

outputCopy
? 2

? 1

? 0

! -1
Note
The polynomial in the first sample is 1000002+x21000002+x^21000002+x2.

The polynomial in the second sample is 1+x21+x^21+x2.

系统生成一个函数f(x)=a0+a1x+a2x2+a3x3+...+a10x10f(x)=a_0+a_1x+a_2x^2+a_3x^3+...+a_{10}x^{10}f(x)=a0+a1x+a2x2+a3x3+...+a10x10，在不多于50次询问的情况下回答一个使f(x)%(1e6+3)=0的x，或者回答-1表示这样的x不存在。
询问：输出一个数x，系统输入f(x)%(1e6+3)的值，需要使用清空缓冲区的相关函数。

线性代数应用题…可以使用矩阵的变换去解n元一次方程组。方法就是把a1−a10a_1-a_{10}a1−a10当做待求量（a0询问一个0直接得到），依次询问1，2，3，…，10（其实只要是十次方不超过longlong的范围都ok，从中选取10个方程，因为有10个待求量）然后就已知了十个方程组成的方程组，把他看成10∗1110*1110∗11的矩阵，进行行变换变换成左侧10∗1010*1010∗10为单位阵的情况，最右侧一列的每一项就是相应的系数，第一项就是a1，以此类推，这样就求得了函数，之后只要把1-1e6+3之间的数都用这个函数试一下就可以了。因为之前询问次数比较少，也可以在本地跑完是0的情况下提交一次验证结果。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=15;
const int mod=1e6+3;
ll qpow(ll a,ll b)
{ll x=a,ans=1;while(b){if(b&1) ans=ans*x%mod;x=x*x%mod;b>>=1;}return ans;
}
ll qqpow(ll a,ll b)
{ll x=a,ans=1;while(b){if(b&1) ans=ans*x;x=x*x;b>>=1;}return ans;
}
ll inv(ll x)
{return qpow(x,mod-2);
}
ll a[N][N];
int main()
{ll a0;printf("? 0\n");fflush(stdout);scanf("%lld",&a0);if(a0==0){printf("! %d",0);return 0;}if(a0==-1) while(1);ll x=(mod-a0)%mod;//ll tmp;for(int i=0;i<10;i++){for(int j=1;j<=10;j++)a[i][j]=qqpow(i+1,j);printf("? %d\n",i+1);fflush(stdout);scanf("%lld",&a[i][11]);if(a[i][11]==0){printf("! %d",i+1);return 0;}if(a[i][11]==-1) while(1);a[i][11]=(a[i][11]-a0+mod)%mod;}
//    for(int i=0;i<10;i++)
//    {//        for(int j=1;j<=11;j++)
//            printf("%11lld ",a[i][j]);
//        printf("\n");
//    }for(int i=1;i<10;i++){for(int k=0;k<i;k++){ll p=a[i][k+1];for(int j=k+1;j<=11;j++){a[i][j]=((a[i][j]-p*a[k][j])%mod+mod)%mod;}
//
//    for(int i=0;i<10;i++)
//    {//        for(int j=1;j<=11;j++)
//            printf("%11lld ",a[i][j]);
//        printf("\n");
//    }}ll q=a[i][i+1];for(int j=i+1;j<=11;j++){a[i][j]=(a[i][j]*inv(q)+mod)%mod;}}
//
//    for(int i=0;i<10;i++)
//    {//        for(int j=1;j<=11;j++)
//            printf("%11lld ",a[i][j]);
//        printf("\n");
//    }for(int i=8;i>=0;i--){for(int k=10;k>i+1;k--){ll p=a[i][k];for(int j=k;j<=11;j++){a[i][j]=((a[i][j]-p*a[k-1][j])%mod+mod)%mod;}//    for(int i=0;i<10;i++)
//    {//        for(int j=1;j<=11;j++)
//            printf("%11lld ",a[i][j]);
//        printf("\n");
//    }}ll q=a[i][i+1];for(int j=1;j<=11;j++){a[i][j]=(a[i][j]*inv(q)+mod)%mod;}}for(int i=0;i<mod;i++)if((qpow(i,1)*a[0][11]%mod+qpow(i,2)*a[1][11]%mod+qpow(i,3)*a[2][11]%mod+qpow(i,4)*a[3][11]%mod+qpow(i,5)*a[4][11]%mod+qpow(i,6)*a[5][11]%mod+qpow(i,7)*a[6][11]%mod+qpow(i,8)*a[7][11]%mod+qpow(i,9)*a[8][11]%mod+qpow(i,10)*a[9][11]%mod)%mod==x){printf("? %d\n",i);fflush(stdout);int tt;scanf("%d",&tt);if(tt==0){printf("! %d",i);return 0;}}printf("! -1");return 0;
}

Educational Codeforces Round 63 (Rated for Div. 2) -E相关推荐

B. Game with Telephone Numbers Educational Codeforces Round 63 (Rated for Div. 2)
题意: 给定字符串,两人游戏轮流删除一个字符,谁先让字符串长度为11且第一个是8即获胜.问先手是否必胜. 思路:0到n-10之间的8的数量大于可移除的数量/2即YES否则NO 参考代码: #inclu ...
Educational Codeforces Round 90 (Rated for Div. 2)（A, B, C, D, E）
Educational Codeforces Round 90 (Rated for Div. 2) Donut Shops 思路分三种情况: a==c/ba == c / ba==c/b这个时候两 ...
Educational Codeforces Round 114 (Rated for Div. 2) （A ~ F）全题解
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划 Educational Codeforces Round 114 (Rated for Div. 2) ...
Educational Codeforces Round 106 (Rated for Div. 2)（A ~ E）题解（每日训练 Day.16 ）
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划目录 Educational Codeforces Round 106 (Rated for Div. ...
Educational Codeforces Round 37 (Rated for Div. 2) 1
Educational Codeforces Round 37 (Rated for Div. 2) A.Water The Garden 题意:Max想给花园浇水.花园可被视为长度为n的花园床,花园 ...
Educational Codeforces Round 89 (Rated for Div. 2)（A, B, C, D）
Educational Codeforces Round 89 (Rated for Div. 2) A. Shovels and Swords 思路题意非常简单,就是得到最多的物品嘛,我们假定a, ...
Educational Codeforces Round 114 (Rated for Div. 2) D. The Strongest Build 暴力 + bfs
传送门文章目录题意: 思路: 题意: 你有nnn个装备槽,每个槽里面有cic_ici个力量加成,对于每个槽只能选一个力量加成,现在给你mmm个力量组合[b1,b2,...,bn][b_1,b_2 ...
Educational Codeforces Round 72 (Rated for Div. 2) D. Coloring Edges dfs树/拓扑找环
传送门文章目录题意: 思路: 题意: 给你一张图,你需要给这个图的边染色,保证如果有环那么这个环内边的颜色不全相同,输出染色方案和用的颜色个数. n,m≤5e3n,m\le5e3n,m≤5e3 思 ...
Educational Codeforces Round 111 (Rated for Div. 2) D. Excellent Arrays 组合数学
传送门文章目录题意: 思路: 题意: 给你一个数组aia_iai,定义一个数组是好的当且仅当对于所有iii都有ai!=ia_i!=iai!=i.定义f(a)f(a)f(a)表示数组aaa中i& ...

Educational Codeforces Round 63 (Rated for Div. 2) -E

Educational Codeforces Round 63 (Rated for Div. 2) -E相关推荐

最新文章

热门文章