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  • Problem D: Integer Inquiry

    Problem D: Integer Inquiry
    Time Limit: 1 Sec Memory Limit: 128 MB
    Submit: 41 Solved: 12
    [Submit][Status][Web Board]
    Description
    One of the first users of BIT’s new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.

    “This supercomputer is great,” remarked Chip. “I only wish Timothy were here to see these results.” (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.)

    Input
    The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative).

    The final input line will contain a single zero on a line by itself.

    Output
    Your program should output the sum of the VeryLongIntegers given in the input.

    Sample Input
    123456789012345678901234567890
    123456789012345678901234567890
    123456789012345678901234567890
    0
    Sample Output
    370370367037037036703703703670
    my answer:

    #include<iostream>
    #include<stdio.h>
    #include<cstring>
    #include<string>
    using namespace std;
    int main()
    {
        char a[101];
        char b[101];
        memset(a,'0',sizeof(a));
        memset(b,'0',sizeof(b));
     
        while(cin>>b)
        {
            int sum[101]={0};
            int t2=strlen(b),p=t2;
            b[t2]='0';
            for(int i=100;t2-1>=0;i--,t2--){
                    b[i]=b[t2-1];
                    b[t2-1]='0';
                }
            while(scanf("%s",a)&&strcmp(a,"0"))
            {
                int t1=strlen(a),q=t1;
                a[t1]='0';
                for(int i=100;t1-1>=0;i--,t1--){
                    a[i]=a[t1-1];
                    a[t1-1]='0';
                }
                for(int j=100;j>=0;j--)
                    b[j]=(b[j]-'0')+(a[j]-'0')+'0';
                for(int i=100;i>=0;i--)
                    sum[i]=b[i]-'0';
     
                for(int j=100;j>=0;j--){
                    sum[j-1]+=sum[j]/10;
                    sum[j]=sum[j]%10;
                }
            }
            memset(a,'0',sizeof(a));
            memset(b,'0',sizeof(b));
            int start=0;
            for(int k=0;k<=100;k++)
                if(sum[k]==0)
                    start++;
                else
                    break;
            for(int k=start;k<=100;k++)
                cout<<sum[k];
            cout<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/NYNU-ACM/p/4236895.html
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