PAT甲级1002
2018-03-08 16:50
399 查看
有两个测试点没有通过
#include <iostream> #include<cstdio> #include<algorithm> #include<string.h> using namespace std; bool COM(int a,int b) { return a>=b; } int main() { int tem_n; double tem_aN; int number=0; double aN[1001]; int n[22]; int K1,K2; memset(aN,0,1001*sizeof(double)); cin>>K1; for(int i=0;i<K1;i++) { cin>>tem_n>>tem_aN; n[i]=tem_n; aN[tem_n]+=tem_aN; number++; } cin>>K2; for(int i=0;i<K2;i++) { cin>>tem_n>>tem_aN; if(aN[tem_n]==0) { aN[tem_n]=tem_aN; n[number++]+=tem_n; } else { n[number]=tem_n; aN[tem_n]+=tem_aN; } if(aN[tem_n]==0) number--; } sort(n,n+number,COM); cout<<number; for(int i=0;i<number;i++) { printf(" %d %.1f",n[i],aN[n[i]]); } return 0; }
相关文章推荐
- PAT甲级1002题解
- PAT 甲级1002_A+B for Polynomials (25)
- PAT甲级1002
- 【PAT】甲级1002 - A+B for Polynomials(多项式加法)
- PAT程序设计练习——甲级1002(两个多项式的解析与合并)
- PAT甲级.1002. A+B for Polynomials (25)
- PAT甲级1002
- 浙大PAT甲级-1002
- PAT-甲级-1002
- PAT 甲级 1002
- PAT-甲级-1002 A+B for Polynomials
- pat甲级1002
- PAT 甲级 1002
- PAT 甲级1002 A+B for Polynomials (25)
- PAT 甲级 1002
- 浙大pat | 浙大pat 牛客网甲级 1002 All Roads Lead to Rome (30)迪杰斯特拉算法改进
- PAT程序设计考题——甲级1002(A+B for Polynomials ) C++实现
- PAT甲级1002 A+B for Polynomials
- PAT甲级1002 A + B
- PAT甲级 1002 A+B for Polynomials (25)