高精度计算：求和、减法与乘法问题-优快云博客

本文链接：https://blog.youkuaiyun.com/Limhhhhh/article/details/50266439

博客主要介绍了高精度计算在解决数学问题中的应用，包括高精度数的加法、减法和乘法。作者在解决ZOJ上的高精度题目时遇到了超时和错误的问题，通过调试和修改代码范围最终解决了问题，分享了这一过程中的困扰与解决方案。

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高精度求和

（应该是zoj上的一道题，但是zoj打不开了）这是一个多个高精度数求和，每个数不超过100位，不超过20个数相加。本来做过两个高精度数相加后做这个题是挺简单的，但是我居然超时都超了好几次，我也觉得很奇怪，数字这么小怎么会超时，结果发现while循环里的输入忘了写结束符EOF，这个简直是郁闷，然后改了又变为WA了，想了很久也没想明白，最后我试着改了一个范围，居然就对了，郁闷，郁闷。。。。。。（必须发个博客调整一下心情）

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    char a[110];
    int b[110],ans[110],N,lenmax=0;
    while(scanf("%d",&N)!=EOF)                                           //第二次没发现这个问题
    {
        memset(ans,0,sizeof(ans));
        for(int i=0;i<N;i++)
        {
            memset(a,0,sizeof(char)*110);
            cin>>a;
            int len=strlen(a);
            if(len>lenmax)
                lenmax=len;
            memset(b,0,sizeof(b));
            for(int j=0;j<len;j++)
            {
                b[j]=a[len-j-1]-'0';
            }
            for(int j=0;j<=lenmax;j++)                                    //当初写的范围是（0，len）想的是一个新的数与前面的和相加最多会进一位，所以算到len就够了，结果一直WA。
            {
                ans[j]+=b[j];
                ans[j+1]+=ans[j]/10;
                ans[j]%=10;
            }
        }
        if(ans[lenmax]!=0)
            lenmax++;
        while(lenmax&&!ans[lenmax])  lenmax--;
        for(int i=lenmax;i>=0;i--)  printf("%d",ans[i]);
        printf("\n");
    }
    return 0;
}

1529: 高精度减法

Time Limit: 1000MS Memory Limit: 65536KB
Total Submit: 425 Accepted: 85 Page View: 1535
Submit Status Discuss

输入多组数据 m n, 计算m-n的值。
m n 均为非负数，长度不超过100位。

2345436 556
100024234535343435436 654654634534653465

2344880
99369579900808781971

注意前导0

SWUST.dreamone

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    char a[105],c[105];
    int b[105],d[105],f[105],h[105],len,len1,len2,res,i,flag=0;
    memset(a,0,sizeof(a));
    memset(c,0,sizeof(c));
    while(scanf("%s%s",a,c)!=EOF)
    {
    len1=strlen(a);
    len2=strlen(c);
    memset(b,0,sizeof(b));
    memset(d,0,sizeof(d));
    memset(f,0,sizeof(f)); 
    for(i=0;i<len1;i++)    
        b[i]=a[len1-i-1]-'0';  
    for(i=0;i<len2;i++)
        d[i]=c[len2-i-1]-'0';          
    while((len1)&&!b[len1-1])    len1--;
    while(len2&&!d[len2-1])  len2--;
    if(len1<len2)   
        flag=1;
    else
        if(len1>len2)
            flag=0;
        else
    for(i=len1-1;i>=0;i--)
    {
        if(b[i]>d[i])
        {
            flag=0;
            break;
        }
        if(b[i]<d[i])
        {
            flag=1;
            break;
        }
    }
    if(flag==1)
    {
        for(i=0;i<len2;i++)
        {
            h[i]=b[i];
            b[i]=d[i];
            d[i]=h[i];
        }
        printf("-");
    }
    res=0;
    len=len1>len2?len1:len2;
    for(i=0;i<=len;i++)
    {
        if(b[i]-d[i]+res>=0)
        {
            f[i]=b[i]-d[i]+res;
            res=0;
        }
        else
        {
            f[i]=b[i]-d[i]+res+10;
            res=-1;
        }
    }
    if(f[len]!=0)
            len+=1;
    while(len&&!f[len])  len--;
    for(i=len;i>=0;i--)  printf("%d",f[i]);
    printf("\n");
    }
    return 0;
}

1019: A * B Problem (Big integer version) 【高精度】

Time Limit: 1000MS Memory Limit: 65536KB
Total Submit: 12 Accepted: 12 Page View: 1451
Submit Status Discuss

计算2个不超过40位的正整数的积.

输入为2行,每一行代表一个数.

输出一行,为2个数的乘积.

11111111111111
1111111111

12345679011110987654321

用数组!

SWUST.Larrous

<span style="font-size:10px;">#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    char a[45],c[45];
    int b[45],d[45],f[85],len,len1,len2,i,j;
    memset(a,0,sizeof(a));
    memset(c,0,sizeof(c));
    while(scanf("%s%s",a,c)!=EOF)
    {
        len1=strlen(a);
        len2=strlen(c);
        memset(b,0,sizeof(b));
        memset(d,0,sizeof(d));
        memset(f,0,sizeof(f));
        for(i=1;i<=len1;i++)
        {
            b[i]=a[len1-i]-'0';
        }
        for(i=1;i<=len2;i++)
        {
            d[i]=c[len2-i]-'0';
        }
        len=len1+len2;
        for(i=1;i<=len2;i++)
        {
            for(j=1;j<=len1;j++)
            {
                f[i+j-1]+=b[j]*d[i];
                f[i+j]+=f[i+j-1]/10;
                f[i+j-1]%=10;
            }
        }
        while(len&&!f[len])  len--;
        for(i=len;i>0;i--)  printf("%d",f[i]);
        printf("\n");
    }
    return 0;
}</span>

1202: A + B Problem (Big integer version) 【高精度】

Time Limit: 1000MS Memory Limit: 65536KB
Total Submit: 5 Accepted: 3 Page View: 2328
Submit Status Discuss

Input and output are the same with problem 1001.
But A and B are big non-negative integers.
The biggest integer is less than 10^500.

Input contains multiple test cases. 
Each case have two big non-negative integers a,b.

For each case,Output a+b.

1234567890987654321
9876543210123456789
123456789123456789
321654987321654987
9081321110693270343633073697474256143563558458718976746753830538032062222085722974121768604305613921745580037409259811952655310075487163797179490457039169594160088430571674960498834085812920457916453747019461644031395307920624947349951053530086146486307198155590763466429392673709525428510973272600608981219760099374675982933766845473509473676470788342281338779191792495900393751209539300628363443012
6538005862664913074813656220643842443844131905754565672075358391135537108795991638155474452610874309742867231360502542308382199053675592825240788613991898567277116881793749340807728335795394301261629479870548736450984003401594705923178314906195914825136973281314862289454100745237769034410057080703111299605127114594552921209928891515242515620324828055912854227507525717981351447473570262981491527798

11111111101111111110
445111776445111776
15619326973358183418446729918118098587407690364473542418829188929167599330881714612277243056916488231488447268769762354261037509129162756622420279071031068161437205312365424301306562421608314759178083226890010380482379311322219653273129368436282061311444171436905625755883493418947294462921030353303720280824887213969228904143695736988751989296795616398194193006699318213881745198683109563609854970810

Note that there are leading zeros!

SWUST.Larrouse

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    char a[505],c[505];
    int b[505],d[505],f[505],len,len1,len2,res,i;
    memset(a,0,sizeof(a));
    memset(c,0,sizeof(c));
    while(scanf("%s%s",a,c)!=EOF)
    {
    len1=strlen(a);
    len2=strlen(c);
    len=len1>len2<span style="font-size:10px;">?len1:len2;
    memset(b,0,sizeof(b));</span>
    memset(d,0,sizeof(d));
    memset(f,0,sizeof(f));
    for(i=0;i<len1;i++)
    { 
        b[i]=a[len1-i-1]-'0';
    }
    for(i=0;i<len2;i++)
    {
        d[i]=c[len2-i-1]-'0';      
    }
    res=0;
    for(i=0;i<=len;i++)
    {
        f[i]=(b[i]+d[i]+res)%10;
        res=(b[i]+d[i]+res)/10;
    }
    if(f[len]!=0)
            len+=1;
    while(len&&!f[len])  len--;
    for(i=len;i>=0;i--)  printf("%d",f[i]);
    printf("\n");
    }
    return 0;
}