HDoj1047Integer Inquiry(大数连续相加)
发布时间：2021-01-02 10:15:29 所属栏目：大数据 来源：网络整理
导读：Integer Inquiry Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19438????Accepted Submission(s): 5087 Problem Description One of the first users of BIT's new supercomputer was Chip D
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Integer InquiryTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19438????Accepted Submission(s): 5087 Problem 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 #include<stdio.h>
#include<string.h>
#define N 1010
void add(int a[],int c[],int len)
{
int f,ans;
for(int i = 0;i < len;i ++){
ans = a[i] + c[i];
f = ans / 10;
c[i+1] += f;
c[i] = ans % 10;
}
}
int main()
{
int T,a[N],c[N],i,j,flag=0;
char str[N];
scanf("%d",&T);
for(j = 1;j <= T;j ++){
memset(a,sizeof(a));
memset(c,sizeof(c));
while(~scanf("%s",str)){
if(str[0]=='0')
break;
flag=1;
int len = strlen(str);
for(i = 0; i < len;i ++)
a[i] = str[len-i-1]-'0';
add(a,c,len);
}
if(c[0]==0&&flag==0)
printf("0n");
else{
for(i = N-1;i > 0;i --){
if(c[i]!=0)
break;
}
for(;i >= 0;i --)
printf("%d",c[i]);
printf("n");
}
if(j!=T)
printf("n");
}
return 0;
}
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