Num 11 : HDOJ: 题目1002 : A+B ( 大数问题 )
发布时间:2020-12-14 02:27:41 所属栏目:大数据 来源:网络整理
导读:原题链接 ? ?? ?? 由于C语言定义的数值型的数据范围有限, ???? 所以不能用简单的加减法来计算数据特别大的数; ???? 这时候就要用到:? 字符数组来进行计算; ???? 字符数组: ???? 通过定义两个数组( 一个为 int 型,一个为 char 型 ); ???? 先将输入的数
原题链接 ????? 由于C语言定义的数值型的数据范围有限, ???? 所以不能用简单的加减法来计算数据特别大的数; ???? 这时候就要用到:? 字符数组来进行计算; ???? 字符数组: ???? 通过定义两个数组( 一个为 int 型,一个为 char 型 ); ???? 先将输入的数字用 char 型数组存储,之后再用语句: ???????? ( int ) a[ i ]= ( char ) b[ i ] - ' 0 ' ; ???? 将输入的数字字符按顺序存入到 int 型数组中( 按 a[ 0 ] 为首位存储 ); ???? 例( 此数组可存储2000位的数 ): #include<stdio.h> #include<string.h> int main() { int i,a[2000],lenb; char b[2000]; scanf("%s",&b); lenb=strlen(b); for(i=0; i<lenb; i++) a[i]=b[i]-'0'; for(i=0; i<lenb; i++) printf("%d",a[i]); } ???? 大数 A+B 的算法: ???? 通过对数组A 和数组 B的每一位相加,并且判断 ???? 若和大于十的时候,和减去十,并且下一位加一即可; 题目:
A + B Problem IITime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 260741????Accepted Submission(s): 50418
Problem Description
I have a very simple problem for you. Given two integers A and B,your job is to calculate the Sum of A + B.
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Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow,each line consists of two positive integers,A and B. Notice that the integers are very large,that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
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Output
For each test case,you should output two lines. The first line is "Case #:",# means the number of the test case. The second line is the an equation "A + B = Sum",Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
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Sample Input
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Sample Output
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AC代码: #include<stdio.h> #include<string.h> #include<stdlib.h> #include<malloc.h> char a[2200],b[2200]; int main() { int p,n,numa[2200],numb[2200]; scanf("%d",&n); for(p=0; p<n; p++) { scanf("%s%s",&a,&b); int i,j,lena,lenb; memset(numa,sizeof(numa));//重置并归零数组!!!!!; memset(numb,sizeof(numb)); lena=strlen(a); lenb=strlen(b); for(j=0,i=lena-1; i>=0; i--) numa[j++]=a[i]-'0';//字符数组 for(j=0,i=lenb-1; i>=0; i--) numb[j++]=b[i]-'0'; for(i=0; i<2200; i++) { numa[i]+=numb[i];//对应位相加; if(numa[i]>=10)//若和超过10,仅保留个位,下一位加一; { numa[i]-=10; numa[i+1]++; } } for(i=2199; ((i>=0)&&numa[i]==0); i--);//除去数组后多余的 '0 ',并且把定位值 i 放在最后( 从后向前输出 )!!!!!; if(i>=0) { printf("Case %d:n",p+1); printf("%s + %s = ",a,b); for(; i>=0; i--) { int q; printf("%d",numa[i]); if(i==0&&p!=n-1) printf("n"); } } printf("n"); } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |