大数阶乘

大数阶乘
时间限制: 3000ms内存限制: 128000KB64位整型: Java 类名:
上一题 提交 运行结果 统计 讨论版 下一题
类型:
没有

添加
题目描述
我们都知道如何计算一个数的阶乘，可是，如果这个数很大呢，我们该如何去计算它并输出它？
输入
输入一个整数m(0

#include<stdio.h>
#include<string.h>
int a[20000];
void gg(int n)
{
    memset(a,0,sizeof(a));
    a[0]=1;
    int i,j;
    for(i=2; i<=n; i++)
    {
        int c=0;
        for(j=0; j<19001; j++)
        {
            int s=a[j]*i+c;
            a[j]=s%10;
            c=s/10;
        }
    }
}
int main()
{
    int n,i,f;
    scanf("%d",&n);
    gg(n);
    for(i=19000;i>=0;i--)
    {
        if(a[i]!=0)
        {
            f=i;
            break;
        }
    }
    for(i=f;i>=0;i--)
        printf(i==0?"%d\n":"%d",a[i]);
    return 0;
}

A+B Problem II
时间限制：3000 ms | 内存限制：65535 KB
难度：3
描述
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
输出
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation.
样例输入
2
1 2
112233445566778899 998877665544332211
样例输出
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

#include<stdio.h>
#include<string.h>
int a2[1005],b2[1005],c[1005];
void gg(char q[2000],int w[2000],int lon)
{
    int i,n=0;
    for(i=lon-1; i>=0; i--)
        w[n++]=q[i]-'0';
}
void ff(int y[2000],int u[2000],int o[2009],int l)
{
    int i;
    for(i=0; i<l; i++)
        o[i]=y[i]+u[i];
    for(i=0; i<l; i++)
    {
        o[i+1]+=o[i]/10;
        o[i]%=10;
    }
}
int main()
{
    int n,an,la,lb,z,k,flag=0;
    scanf("%d",&n);
    an=n;
    while(n--)
    {
        if(flag==1)
            printf("\n");
        char a1[1005],b1[1005];
        scanf("%s%s",a1,b1);
        memset(a2,0,sizeof(a2));
        memset(b2,0,sizeof(b2));
        memset(c,0,sizeof(c));
        la=strlen(a1);
        lb=strlen(b1);
        gg(a1,a2,la);
        gg(b1,b2,lb);
        if(la>lb)
            la^=lb^=la^=lb;
        ff(a2,b2,c,lb);
        printf("Case %d:\n",an-n);
        printf("%s + %s = ",a1,b1);
        int d;
        if(c[lb])
            d=lb;
        else
            d=lb-1;
        for(int i=d; i>=0; i--)
            printf(i==0?"%d\n":"%d",c[i]);
        flag=1;
    }
    return 0;
}