摘要：Problem DescriptionA Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)Your task is to take a number as input, and print that Fibonacci number 阅读全文

摘要：Problem DescriptionFibonacci数列，定义如下：f(1)=f(2)=1f(n)=f(n-1)+f(n-2) n>=3。计算第n项Fibonacci数值。Input输入第一行为一个整数N，接下来N行为整数Pi（1#includeint m=10000;struct bignum{ int s[100]; int l;}f[1001];bignum operator+(bignum a,bignum b){ int i,d=0; if(b.l0;i--) printf("%04d",a.s[i]); printf("\n");} 阅读全文

摘要：Problem DescriptionGiven an integer N(0 ≤ N ≤ 10000), your task is to calculate N!InputOne N in one line, process to the end of file.OutputFor each N, output N! in one line.Sample Input123Sample Output126高精度运算#include#includeint m=10000;struct bignum{ int s[20001];//存放结果 int l;//数组的长度}a;bignum... 阅读全文

摘要：题目描述：输入两个正整数A,B,计算A * B输出结果输入：输入包含多行，第一行表示测试数目,以后每行包含两个正整数A,B(A,B的位数小于200位)输出：针对每一行输入，输出一行代表A*B的结果此题是一个高精度问题，采取的算法是模拟乘法过程。用到字符三个数组。s1,s2用来接收A，Ｂ的值，S3用来存取结果。S3从数组下标最大开始存。代码如下：#include#include#includechar s1[201],s2[201],s3[401];int main(){ int i,j,n,k1,k2,k,t,d,m; scanf("%d",&n); while(n 阅读全文

摘要：此题的思路是模拟加法运算的过程，首先用s1,s2两个字符串数组接受两个大数。再将其转化为数字。s3数组保存加法之后的数字。注意输出时除去前导0 ，代码如下：#include#includeint main(){ void zhuan(int ,int ,char s1[],char s2[],char s[]); char s1[1001],s2[1001],s[1001]; int i,a,b,t,j; memset(s1,0,sizeof(s1)); memset(s2,0,sizeof(s2)); scanf("%d",&t);getchar(); ... 阅读全文