题目链接(POJ):?id=1503
题目链接(HDOJ):?pid=1047
Integer Inquiry
Time Limit:1000MSMemory Limit:10000K
Total Submissions:30856Accepted:12007
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
1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678900
Sample Output
370370367037037036703703703670
Source
题意:求n个大数之和
题解; 大数加法模板题——POJ和HDU上格式出处的要求稍有不同——HDOJ上的描述好混乱。
AC代码(POJ)
#include<iostream>#include<cstring>#include<string>#define maxn 300using namespace std;int numx[maxn],numy[maxn],n;string str,tmp;string Add(string x,string y){string res="";memset(numx,0,sizeof(numx));memset(numy,0,sizeof(numy));int lenx=x.size(),leny=y.size();int maxlen=lenx>leny ? lenx:leny;for(int i=0;i<lenx;i++)numx[lenx-i-1]=x[i]-'0';for(int i=0;i<leny;i++)numy[leny-i-1]=y[i]-'0';for(int i=0;i<=maxlen;i++){numx[i]+=numy[i];if(numx[i]>9){numx[i+1]+=numx[i]/10;numx[i]%=10;}}int i=maxlen+2;for(;i>0&&!numx[i];)i–;for(;i>=0;i–)res+=numx[i]+'0';return res;}int main(){string sum="0";while(cin>>str&&str!="0")sum=Add(sum,str);cout<<sum<<endl;return 0;}
AC代码(HDU):
#include<iostream>#include<cstring>#include<string>#define maxn 300using namespace std;int numx[maxn],numy[maxn],n;string str,tmp;string Add(string x,string y){string res="";memset(numx,0,sizeof(numx));memset(numy,0,sizeof(numy));int lenx=x.size(),leny=y.size();int maxlen=lenx>leny ? lenx:leny;for(int i=0;i<lenx;i++)numx[lenx-i-1]=x[i]-'0';for(int i=0;i<leny;i++)numy[leny-i-1]=y[i]-'0';for(int i=0;i<=maxlen;i++){numx[i]+=numy[i];if(numx[i]>9){numx[i+1]+=numx[i]/10;numx[i]%=10;}}int i=maxlen+2;for(;i>0&&!numx[i];)i–;for(;i>=0;i–)res+=numx[i]+'0';return res;}int main(){while(cin>>n){while(n–){string sum="0";while(cin>>str&&str!="0"){sum=Add(sum,str);}cout<<sum<<endl;if(n)cout<<endl;}}return 0;}
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作者:MummyDing
出处:
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