POJ 1503(高精度整数加法) 解题报告

/*_____________________________________POJ 1503题_________________________________________________ Integer Inquiry Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 18770 Accepted: 7384 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 _________________________________________________________________________________________________*/ #include<stdio.h> #include<string.h> void Convert(char p[],int n) //翻转 { char temp; int i,j; for(i=0,j=strlen(p)-1; i<j; i++,j--) { temp=p[i]; p[i]=p[j]; p[j]=temp; } } int main() { int i,j,k,carry,remain; char a[110]={'/0'}, b[110]={'/0'}, sum[110]={'/0'}; //由于相加后可能超过100位所以长度设为110位 //FILE *fin=fopen("input.txt","r"); //fscanf(fin,"%s",a); scanf("%s",a); Convert(a,strlen(a)); //将a[]翻转为从低位到高位存储 //fscanf(fin,"%s",b); scanf("%s",b); while(!(strcmp(b,"0")==0)) //输入未结束 { Convert(b,strlen(b)); //将b[]翻转为从低位到高位存储 for(carry=0,k=0,i=0,j=0; i<strlen(a)&&j<strlen(b); i++,j++) { remain=(a[i]-'0'+b[j]-'0'+carry)%10; //余数 carry=(a[i]-'0'+b[j]-'0'+carry)/10; //进位,为0或1 sum[k++]=remain+'0'; //和sum[]也是从低位到高位存放 } while(i<strlen(a)) //串a[]的位数多 { remain=(a[i]-'0'+carry)%10; carry=(a[i]-'0'+carry)/10; sum[k++]=remain+'0'; i++; } while(j<strlen(b)) //串b[]的位数多 { remain=(b[j]-'0'+carry)%10; carry=(b[j]-'0'+carry)/10; sum[k++]=remain+'0'; j++; } if(carry) //a[],b[]的位数相同，但最高位还有进位,但加法的进位只可能是一位 sum[k++]=carry+'0'; for(i=0;i<110;i++) //将a[],b[]清空 a[i]=b[i]='/0'; strcpy(a,sum); //将sum[]拷贝给a[],做下一次加法的一个加数 //fscanf(fin,"%s",b); //b[]是另一个加数 scanf("%s",b); } Convert(sum,strlen(sum)); printf("%s/n",sum); return 0; } /****************************************************************************************** 第一次没考虑到的情况： (1)a，b位数同时用尽时可能还有剩的进位carry，导致9+9结果为8 (2)99+9的情况，a的位数用尽后还有进位carry，总之就是没考虑最高位进位 所以应该加上if(carry) {sum[k++]=carry+'0';} ******************************************************************************************/