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.
Sample Input
2 1 2.4 0 3.2
2 2 1.5 1 0.5
Sample Output
3 2 1.5 1 2.9 0 3.2
//本算法利用同一数组进行计算,较简便,满分
#include <stdio.h>
#define N 1001
int main(){
float a[N] = { 0 };
int k, e;
float c;
int cnt = 0;
scanf("%d", &k);
for (int i = 0; i < k; i++){
scanf("%d %f", &e, &c);
a[e] += c;
}
scanf("%d", &k);
for (int i = 0; i < k; i++){
scanf("%d %f", &e, &c);
a[e] += c;
}
for (int i = 0; i < N; i++){
if (a[i] != 0){
cnt++;
}
}
printf("%d", cnt);
for (int i = N-1; i >= 0; i--){
if (a[i] != 0){
printf(" %d %.1f", i, a[i]); //不必在最后一个数后加回车
}
}
return 0;
}
//本算法利用三个数组,较繁琐,且有漏洞(哪位大神不吝指教,不胜感激~),23分
#include <stdio.h>
#define N 1001
int main(){
int k1;
scanf("%d", &k1);
float a[N] = { 0 }, b[N] = { 0 };
for (int i = 0; i < 2 * k1; i++){
scanf("%f", &a[i]);
}
int k2;
scanf("%d", &k2);
for (int i = 0; i < 2 * k2; i++){
scanf("%f", &b[i]);
}
float c[N] = { 0 };
int i = 0, j = 0, k = 0;
while (i < N - 1){
if (a[i] == b[j]){
if (a[i + 1] != 0 || b[j + 1] != 0){
c[k] = a[i];
c[k + 1] = a[i + 1] + b[j + 1];
if (c[k + 1] != 0){
k += 2;
}
i += 2;
j += 2;
}
else{
break;
}
}
else{
if (a[i] < b[j]){
c[k] = b[j];
c[k + 1] = b[j + 1];
j += 2;
}
else{
c[k] = a[i];
c[k + 1] = a[i + 1];
i += 2;
}
k += 2;
}
}
printf("%d ", k/2);
for (int t = 0; t < k-1; t++){
if (t % 2 == 0){
printf("%d ", (int)c[t]);
}
else{
printf("%.1f ", c[t]);
}
}
printf("%.1f\n", c[k-1]); //最后一行加回车
return 0;
}