高精度

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

Description

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。

Input

Output

• Sample Input
• Raw

2345436 556100024234535343435436 654654634534653465

• Sample Output
• Raw

234488099369579900808781971

Hint

注意前导0

Source

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

Description

计算2个不超过40位的正整数的积.

Input

输入为2行,每一行代表一个数.

Output

输出一行,为2个数的乘积.

• Sample Input
• Raw

111111111111111111111111

• Sample Output
• Raw

12345679011110987654321

Hint

用数组!

Source

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

Description

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

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

Output

For each case,Output a+b.

• Sample Input
• Raw

1234567890987654321987654321012345678912345678912345678932165498732165498790813211106932703436330736974742561435635584587189767467538305380320622220857229741217686043056139217455800374092598119526553100754871637971794904570391695941600884305716749604988340858129204579164537470194616440313953079206249473499510535300861464863071981555907634664293926737095254285109732726006089812197600993746759829337668454735094736764707883422813387791917924959003937512095393006283634430126538005862664913074813656220643842443844131905754565672075358391135537108795991638155474452610874309742867231360502542308382199053675592825240788613991898567277116881793749340807728335795394301261629479870548736450984003401594705923178314906195914825136973281314862289454100745237769034410057080703111299605127114594552921209928891515242515620324828055912854227507525717981351447473570262981491527798

• Sample Output
• Raw

1111111110111111111044511177644511177615619326973358183418446729918118098587407690364473542418829188929167599330881714612277243056916488231488447268769762354261037509129162756622420279071031068161437205312365424301306562421608314759178083226890010380482379311322219653273129368436282061311444171436905625755883493418947294462921030353303720280824887213969228904143695736988751989296795616398194193006699318213881745198683109563609854970810

Hint

Note that there are leading zeros!

Source

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;}