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            posts - 21, comments - 2, trackbacks - 0, articles - 0

            ZOJ1292 , HDU 1047 Integer Inquiry

            Posted on 2011-09-05 23:22 acpeng 閱讀(773) 評論(0)  編輯 收藏 引用 所屬分類: ACM程序
            大數(shù)相加,字符模擬
            題目鏈接:
            ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1292
            HDU:http://acm.hdu.edu.cn/showproblem.php?pid=1047

            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.


            This problem contains multiple test cases!

            The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

            The output format consists of N output blocks. There is a blank line between output blocks.


            Sample Input

            1

            123456789012345678901234567890
            123456789012345678901234567890
            123456789012345678901234567890
            0


            Sample Output

            370370367037037036703703703670

            AC代碼:

            /*
            Accepted 1047 0MS 188K 1256 B C++ 
            */

            #include
            <stdio.h>
            #include
            <string.h>
            #define MaxSize 1000
            void FunBigAdd(char *str1,char *str2);
            int main()
            {
                
            char IntStr[MaxSize]="\0",str[MaxSize]="\0";
                
            int T,i;
                scanf(
            "%d",&T);
                
            while(T--)
                
            {
                    memset(str,
            0,MaxSize);
                    scanf(
            "%s",IntStr);
                    
            while(strcmp(IntStr,"0")!=0)
                    
            {
                        FunBigAdd(str,IntStr);
                        memset(IntStr,
            0,MaxSize);
                        scanf(
            "%s",IntStr);
                    }

                    i
            =0;
                    
            while(str[i]=='0' && str[i]!='\0')i++;
                    
            if(str[i]=='\0')printf("0\n");
                    
            else printf("%s\n",str+i);
                    
            if(T!=0)printf("\n");
                }

                
            return 0;
            }

            void strRevg(char *str)
            {
                
            int i;char temp;
                
            int lgth=(int)strlen(str)-1;
                
            for(i=0;i<lgth-i;i++)
                
            {
                    temp
            =*(str+i);
                    
            *(str+i)=*(str+lgth-i);
                    
            *(str+lgth-i)=temp;
                }

            }

            void FunBigAdd(char *str1,char *str2)
            {    
                
            int i=0,dec=0,tmp;
                strRevg(str1);
                strRevg(str2);
                
            while(*(str1+i)!='\0' || *(str2+i)!='\0')
                
            {
                    
            if(*(str1+i)=='\0')
                    
            {
                        
            *(str1+i)='0';
                        
            *(str1+i+1)='\0';
                    }

                    
            if(*(str2+i)=='\0')
                    
            {
                        
            *(str2+i)='0';
                        
            *(str2+i+1)='\0';
                    }

                    tmp
            =*(str1+i)-'0'+(*(str2+i)-'0')+dec;
                    
            *(str1+i)=tmp%10+'0';
                    dec
            =tmp/10;
                    i
            ++;
                }

                
            if(dec!=0*(str1+(i++))=dec+'0';
                
            *(str1+i)='\0';
                strRevg(str1);
            }



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