1002. A+B for Polynomials(25)—PAT 甲级

2018/04/29 16:51
阅读数 15

<font size=4 color=#87CEFA>This time,you are supposed to find A+B where A+B are two polynomials.</font> <br> ##<font color=#FFA500>Input</font>## <font color=#87CEFA size=4 style="line-height:1.7;">Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a
polynomial: K N1 aN1 N2 aN2 ... Nk aNK, Where k is the number of nonezero terms in the polynomial, Ni and aNi (i = 1 , 2,..., k) are the exponents and coefficients, respectively . It is given that 1< = K < = 10,0< = NK < ... < N2 < N1 <= 1000. </font> ##<font color=#FFA500>Output</font>## <font color=#87CEFA size=4 style="line-height:1.7;">For each test case you should output the sum of A and B in one line, with the same format asthe input . Notice that there must be No extra space at the end of each line. Please be accurate to 1 decimal place. </font> ##<font color=#FFA500>Sample Input</font>## <font color=#87CEFA size=4 style="line-height:1.7;">2 1 2.4 0 3.2 2 2 1.5 1 0.5 </font> ##<font color=#FFA500>Sample Output</font>## <font color=#87CEFA size=4 style="line-height:1.7;">3 2 1.5 1 2.9 0 3.2 </font> <font color=#20B2AA size=4 style="line-height:1.7;">题目大意:多项式合并同类项,按输入格式输出最后多项式的项数、各项的指数和系数。</font> <font color=#F08080 size=4 style="line -height:1.7;">分析:构造一个整型数组存放多项式的系数,输入时将相同指数的系数累加,如果累加之前数组存放的值为0,那么多项式的项数加1;如果累加之后等于0,那么多项式的项数减一。注意:多项式的指数都是整数,不要瞎想了~ </font>

#include <iostream>
using namespace std;

double s[1005]={0};
int main() {
    int k,n,time=2,count=0; 
    double ak;
    while(time--){
        scanf("%d",&k);
        while(k--){
            scanf(" %d %lf",&n,&ak);
            if(s[n]==0)count++;
            s[n]+=ak;
            if(s[n]==0)count--;
        }
    }
    printf("%d",count);
    for(int i=1005;i>=0;i--){
        if(s[i]!=0){
            printf(" %d %.1lf",i,s[i]);
        }
    }
    printf("\n");
    return 0;
}
展开阅读全文
打赏
0
0 收藏
分享
加载中
更多评论
打赏
0 评论
0 收藏
0
分享
返回顶部
顶部