[{"createTime":1735734952000,"id":1,"img":"hwy_ms_500_252.jpeg","link":"https://activity.huaweicloud.com/cps.html?fromacct=261f35b6-af54-4511-a2ca-910fa15905d1&utm_source=V1g3MDY4NTY=&utm_medium=cps&utm_campaign=201905","name":"华为云秒杀","status":9,"txt":"华为云38元秒杀","type":1,"updateTime":1735747411000,"userId":3},{"createTime":1736173885000,"id":2,"img":"txy_480_300.png","link":"https://cloud.tencent.com/act/cps/redirect?redirect=1077&cps_key=edb15096bfff75effaaa8c8bb66138bd&from=console","name":"腾讯云秒杀","status":9,"txt":"腾讯云限量秒杀","type":1,"updateTime":1736173885000,"userId":3},{"createTime":1736177492000,"id":3,"img":"aly_251_140.png","link":"https://www.aliyun.com/minisite/goods?userCode=pwp8kmv3","memo":"","name":"阿里云","status":9,"txt":"阿里云2折起","type":1,"updateTime":1736177492000,"userId":3},{"createTime":1735660800000,"id":4,"img":"vultr_560_300.png","link":"https://www.vultr.com/?ref=9603742-8H","name":"Vultr","status":9,"txt":"Vultr送$100","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":5,"img":"jdy_663_320.jpg","link":"https://3.cn/2ay1-e5t","name":"京东云","status":9,"txt":"京东云特惠专区","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":6,"img":"new_ads.png","link":"https://www.iodraw.com/ads","name":"发布广告","status":9,"txt":"发布广告","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":7,"img":"yun_910_50.png","link":"https://activity.huaweicloud.com/discount_area_v5/index.html?fromacct=261f35b6-af54-4511-a2ca-910fa15905d1&utm_source=aXhpYW95YW5nOA===&utm_medium=cps&utm_campaign=201905","name":"底部","status":9,"txt":"高性能云服务器2折起","type":2,"updateTime":1735660800000,"userId":3}]
用泰勒展开式求sinx近似值的多项式为:
输入x求sinx的近似值,要求误差不大于0.00001。
<>输入格式:
直接输入一个实型数据。没有其它任何附加字符。
<>输出格式:
直接输出保留3位小数的实型结果。
<>输入样例:
2.5
<>输出样例:
0.598
Note
:这一题隐藏了一些小小的知识点,所以特此记录,还是蛮有意思的一道题目,看注释就能看懂。去搜了一些博客告诉我,fabs的参数为double型,返回值也是double型而abs的参数为int型,返回值也是int型,abs是求一个整数的绝对值,而fabs是求一个实数的绝对值。。所以我以为WA的点是这里。但其实dubug一下之后发现是次数的问题。power需要开到很大才能给到十万分之一的精确度,至于绝对值那儿,abs和fabs都能AC,既然是实数的比较那还是用fabs吧,免得以后在其他题目上出了问题。
#include <iostream> #include <cstdio> #include <cmath> using namespace std; int
main(void) { double x, jinsi = 0; // 定义输入的x和近似值 long long power = 1, xishu = 1,
fenmu= 1; // 次数,交错的系数和分母的阶乘 cin >> x; while (fabs(jinsi - sin(x)) > 1e-5) //
0.00001可用1e-5替换 { jinsi += xishu * pow(x, power) / fenmu; xishu *= -1; //
一加一减,交错系数 power += 2; // 次数+2 fenmu *= (power - 1) * power; //
分母阶乘不要重新计算!比如5!到7!直接乘6和7就好了 } printf("%.3lf", jinsi); }
自己试着不要看把代码码出来哦,测试样例可以直接复制,加油小伙汁小改改,为了更美好的未来呢~~~