原題目:

源代碼:
#include<iostream>
#include<vector>
#include<stdio.h>
#include<cmath>
using namespace std;
struct Ploy{
int n;
float a;
};
float res[1000005];
vector<Ploy> x, y;
int main(){
int k1, k2;
cin>>k1;
for(int j = 0; j < k1; j ++){
Ploy tmp;
cin>>tmp.n>>tmp.a;
x.push_back(tmp);
}
cin>>k2;
for(int j = 0; j < k2; j ++){
Ploy tmp;
cin>>tmp.n>>tmp.a;
y.push_back(tmp);
}
for(int i = 0; i < k1; i ++){
int e = x[i].n;
float f = x[i].a;
for(int j = 0; j < k2; j ++){
res[y[j].n + e] += (f * y[j].a);
}
}
int k_res = 0, max_i;
for(int i = 0; i < 1000005; i ++){
if(res[i] != 0){
max_i = i;
k_res ++;
}
}
cout<<k_res<<" ";
int flag = 0;
for(int i = max_i; i >= 0; i --){
if(res[i] != 0){
cout<<i<<" ";
printf("%.1f", res[i]);
flag ++;
if(flag != k_res)
cout<<" ";
}
}
return 0;
}
已AC:
易失分點:
規模較小,直接用數組存結果沒問題
如果用連結清單存結果,記得删掉值為0的節點
(題目樣例比較簡單,其實系數為浮點型時無法用是否等于0判斷)