PAT 1002
2017-02-26 13:51
155 查看
水题#include<stdio.h>
#include<string.h>
int main()
{
int k, i, p,num;
double a[1005] = {0}, q;
scanf("%d", &k);
for (i = 0; i < k; i++) {
scanf("%d", &p);
scanf("%lf", &q);
a[p] += q;
}
scanf("%d", &k);
for (i = 0; i < k; i++) {
scanf("%d", &p);
scanf("%lf", &q);
if (a[p] != 0)
a[p] += q;
else {
a[p] = 0;
a[p] += q;
}
}
num = 0;
for (i = 1004; i >= 0; i--) {
if (a[i] != 0)
num++;
}
printf("%d", num);
for (i = 1004; i >= 0; i--) {
if (a[i] != 0)
printf(" %d %.1lf", i,a[i]);
}
printf("\n");
return 0;
}
#include<string.h>
int main()
{
int k, i, p,num;
double a[1005] = {0}, q;
scanf("%d", &k);
for (i = 0; i < k; i++) {
scanf("%d", &p);
scanf("%lf", &q);
a[p] += q;
}
scanf("%d", &k);
for (i = 0; i < k; i++) {
scanf("%d", &p);
scanf("%lf", &q);
if (a[p] != 0)
a[p] += q;
else {
a[p] = 0;
a[p] += q;
}
}
num = 0;
for (i = 1004; i >= 0; i--) {
if (a[i] != 0)
num++;
}
printf("%d", num);
for (i = 1004; i >= 0; i--) {
if (a[i] != 0)
printf(" %d %.1lf", i,a[i]);
}
printf("\n");
return 0;
}
相关文章推荐
- PAT1002 写出这个数
- PAT甲级1002
- PAT(乙级)1002
- pat 甲1002. A+B for Polynomials(map)
- PAT 乙级 1002. 写出这个数 (20)
- 1002 写出这个数 PAT
- PAT1002题解
- pat--1002
- PAT乙级1002. 写出这个数 (20)
- pat乙级1002. 写出这个数 (20)
- 【PAT】A1002. A+B for Polynomials(输出格式问题、sort)
- DAY 2 PAT 1002. 写出这个数 (20)
- PAT 1002
- PAT_1002: A+B for Polynomials
- PAT 1002. A+B for Polynomials C语言
- PAT上机题解析之1002. 写出这个数 (20)
- PAT-Advanced Level- 1002 两个多项式相加
- PAT (Basic Level) Practise (中文)1002. 写出这个数 (20)
- PAT 乙级题:1002. 写出这个数 (20)
- PAT乙级1002