hdu 5095 多项式模拟+有坑

http://acm.hdu.edu.cn/showproblem.php?pid=5095

就是把ax^2 + by^2 + cy^2 + dxy + eyz + fzx + gx + hy + iz + j转换成ap + bq + cr + du + ev + fw + gx + hy + iz + j

注意的地方就是开头不要有+，系数为0不输出，系数为1的话不能输出系数除非指数为0等等

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <cstring>

#include <string>

#include <queue>

#include <map>

#include <iostream>

#include <algorithm>

using namespace std;

#define RD(x) scanf("%d",&x)

#define RD2(x,y) scanf("%d%d",&x,&y)

#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define clr0(x) memset(x,0,sizeof(x))

#define clr1(x) memset(x,-1,sizeof(x))

#define eps 1e-9

const double pi = acos(-1.0);

typedef long long LL;

typedef unsigned long long ULL;

const int modo = 1e9 + 7;

const int INF = 0x3f3f3f3f;

const int inf = 0x3fffffff;

const LL _inf = 1e18;

const int maxn = 105,maxm = 10005;

int p[maxn],n,k;

char q[maxn] = {'p','q','r','u','v','w','x','y','z','\n'};

char ch[maxn];

void work()

{

    for(int i = 0;i < 10;++i){

        RD(p[i]);

//        scanf("%s",ch);

//        if(strlen(ch) > 10)

//            while(1);

//        sscanf(ch,"%d",&p[i]);

    }

    int k = 0;

    while(k < 10 && p[k] == 0)k++;

    if(k == 10){

        puts("0");

        return ;

    }

    if(k == 9){

        printf("%d\n",p[9]);

        return;

    }

    if(p[k] == 1)

        printf("%c",q[k]);

    else if(p[k] == -1)

        printf("-%c",q[k]);

    else

        printf("%d%c",p[k],q[k]);

    for(int i = k+1;i < 9;++i){

        if(p[i] > 1)

            printf("+%d%c",p[i],q[i]);

        else if(p[i] < -1)

            printf("%d%c",p[i],q[i]);

        else if(p[i] == 1)

            printf("+%c",q[i]);

        else if(p[i] == -1)

            printf("-%c",q[i]);

    }

    if(p[9] == 0)

        puts("");

    else if(p[9] > 0)

        printf("+%d\n",p[9]);

    else

        printf("%d\n",p[9]);

    return;

}

int main()

{

    int _;RD(_);

    while(_--){

        work();

    }

    return 0;

}