1002. A+B for Polynomials
来源:互联网 发布:网络写手如何赚钱 编辑:程序博客网 时间:2024/04/30 14:39
This time, you are supposed to find A+B where A and B are two polynomials.
Input
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 nonzero 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.
Output
For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.
#include <stdio.h>#include <stdlib.h>#define N 1000 int main (){ int num,i,index; float rate; float ratio[N+1]={0.0f}; scanf("%d",&num); for (i=0;i<num;i++) { scanf("%d %f",&index,&rate); ratio[index]+=rate; } scanf("%d",&num); for (i=0;i<num;i++) { scanf("%d %f",&index,&rate); ratio[index]+=rate; } int count=0; for (i=0;i<=N;i++) { if(ratio[i]!=0.0f) count++; } printf("%d",count); for( i=N;i>=0;i--) { if(ratio[i]!=0.0f) printf(" %d %.1f",i,ratio[i]); } printf("\n"); system("pause"); return 0; }
0 0
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002.A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- 1002. A+B for Polynomials
- git安装
- Java不同压缩算法的性能比较
- 将layout文件转化成View对象的方法
- Deep Learning(深度学习)学习笔记整理系列之(七)
- Hive创建索引
- 1002. A+B for Polynomials
- NYOJ 题目927 The partial sum problem(dfs,剪枝)
- openMPM源码分析(四)
- Deep Learning(深度学习)学习笔记整理系列之(八)
- stack栈
- 关于文件读取的第一次尝试
- try{}catch{} 快捷键
- Hadoop之表的关联
- Android反编译-反编译工具和方法