Sichuan 2017 FSimple Algebra(打表）

F. Simple Algebra
Given function f(x, y) = ax2 + bxy + cy2
, check if f(x, y) ≥ 0 holds for all x, y ∈ R.
Input
The input contains zero or more test cases and is terminated by end-of-file.
Each test case contains three integers a, b, c.
• −10 ≤ a, b, c ≤ 10
• The number of tests cases does not exceed 104
.
Output
For each case, output “Yes” if f(x, y) ≥ 0 always holds. Otherwise, output “No”.
Sample Input
1 -2 1
1 -2 0
0 0 0
Sample Output
Yes
No
Yes
题意：对于输入 的a，b,c,是否对于任意的x，y(x,y∈R）均满足f（x，y)>=0,满足输出Yes，否则输出No。
题解：打表，本地check完，将表交上去。
打表主要注意的几点：
1：表一定要打完（即后面要return 0，判断是否打完）
2：每个输出后加一个“,”,方便数组赋值。
3：数组大小要开好，计算方便。
打表：

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<vector>#include<queue>#include<algorithm>#include<map>using namespace std;typedef long long int ll;typedef pair<int,int>pa;const int N=2e5+10;const int MOD=1e9+7;const ll INF=1e18;int read(){    int x=0;    char ch = getchar();    while('0'>ch||ch>'9')ch=getchar();    while('0'<=ch&&ch<='9')    {        x=(x<<3)+(x<<1)+ch-'0';        ch=getchar();    }    return x;}/************************************************************/int solve(int a,int b,int c){    for(int i=-1000;i<=1000;i++)    {        for(int j=-1000;j<=1000;j++)        {            if(a*i*i+b*i*j+c*j*j<0)            {                return 0;            }        }    }    return 1;}int main(){   // freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);    for(int a=-10;a<=10;a++)    {        for(int b=-10;b<=10;b++)        {            for(int c=-10;c<=10;c++)            {                if(solve(a,b,c)==1)                    printf("1,");                    else                    printf("0,");            }        }    }      return 0;}

代码：

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<vector>#include<queue>#include<algorithm>#include<map>using namespace std;typedef long long int ll;typedef pair<int,int>pa;const int N=2e5+10;const int MOD=1e9+7;const ll INF=1e18;int read(){    int x=0;    char ch = getchar();    while('0'>ch||ch>'9')ch=getchar();    while('0'<=ch&&ch<='9')    {        x=(x<<3)+(x<<1)+ch-'0';        ch=getchar();    }    return x;}int x,y,z;/************************************************************/int ans[21][21][21]={"将打好的表存到这"};int main(){    while(~scanf("%d%d%d",&x,&y,&z))    {        if(ans[x+10][y+10][z+10]==1)            puts("Yes");        else            puts("No");    }}