I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
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.
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.
Sample Input
2 1 2 112233445566778899 998877665544332211
Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
AC代码：
#include<iostream> #include<string.h> using namespace std; #include<stdio.h> #define MAX 1010 int add1[MAX], add2[MAX], res[MAX]; char tmp1[MAX], tmp2[MAX]; int main() { int N, i, j, len, len1, len2, tmp, k; scanf("%d",&N); getchar(); for(j=0;j<N;j++) { memset(add1,0,sizeof(add1)); memset(add2,0,sizeof(add2)); memset(res,0,sizeof(res)); memset(tmp1,0,sizeof(tmp1)); memset(tmp2,0,sizeof(tmp2)); scanf("%s %s",tmp1,tmp2); len1 = strlen(tmp1); len2 = strlen(tmp2); for(i=len1-1,k=0;i>=0;i--) add1[k++] = tmp1[i] - '0'; for(i=len2-1,k=0;i>=0;i--) add2[k++] = tmp2[i] - '0'; tmp = 0; if(len1 >= len2) { for(i=0;i<=len1;i++) { res[i] = (add1[i] + add2[i] +tmp)%10; tmp = (add1[i] + add2[i] +tmp)/10; } } else if(len1 < len2) { for(i=0;i<=len2;i++) { res[i] = (add1[i] + add2[i] +tmp)%10; tmp = (add1[i] + add2[i] +tmp)/10; } } if(len1 >= len2) len = len1; else len = len2; printf("Case %d: %s + %s = ",j+1, tmp1 , tmp2); if(res[len]!=0) printf("%d",res[len]); for(i=len-1;i>=0;i--) printf("%d",res[i]); printf(" "); if(j!=N-1) printf(" "); } return 0; }
my AC:
#include<iostream> #include<stdio.h> #include<string.h> #define N 1010 using namespace std; char a[N],b[N]; int c[N],d[N],ans[N]; int main(){ int t ; cin>>t; getchar(); for(int i=1;i<=t;i++) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); memset(d,0,sizeof(d)); memset(ans,0,sizeof(ans)); cin>>a>>b; int la=strlen(a); int lb=strlen(b); for(int i=0;i<la;i++) c[i]=a[la-1-i]-'0'; for(int i=0;i<lb;i++) d[i]=b[lb-1-i]-'0'; int lans; (la>lb)?lans=la:lans=lb; int temp=0; for(int i=0;i<=lans;i++){ ans[i]=(c[i]+d[i]+temp)%10; temp=(c[i]+d[i]+temp)/10; } cout<<"Case "<<i<<":"<<endl; cout<<a<<" + "<<b<<" = "; for(int i=lans-1;i>=0;i--) cout<<ans[i]; cout<<endl; if(i!=t)cout<<endl; } }
#include<iostream> #include<stdio.h> #include<string.h> #define N 1010 using namespace std; char a[N],b[N]; int c[N],d[N],ans[N]; int main(){ int t ; cin>>t; getchar(); for(int i=1;i<=t;i++) { fill(a,a+N,0); fill(b,b+N,0); fill(c,c+N,0); fill(d,d+N,0); fill(ans,ans+N,0); cin>>a>>b; int la=strlen(a); int lb=strlen(b); for(int i=0;i<la;i++) c[i]=a[la-1-i]-'0'; for(int i=0;i<lb;i++) d[i]=b[lb-1-i]-'0'; int lans; (la>lb)?lans=la:lans=lb; int temp=0; for(int i=0;i<=lans;i++){ ans[i]=(c[i]+d[i]+temp)%10; temp=(c[i]+d[i]+temp)/10; } cout<<"Case "<<i<<":"<<endl; cout<<a<<" + "<<b<<" = "; for(int i=lans-1;i>=0;i--) cout<<ans[i]; cout<<endl; if(i!=t)cout<<endl; } }
#include<iostream> #include<string.h> using namespace std; #include<stdio.h> #define MAX 1010 int add1[MAX], add2[MAX], res[MAX]; char tmp1[MAX], tmp2[MAX]; int main() { int N, i, j, len, len1, len2, tmp, k; scanf("%d",&N); getchar(); for(j=0;j<N;j++) { memset(add1,0,sizeof(add1)) ; ||用0填充 memset(add2,0,sizeof(add2)); memset(res,0,sizeof(res)); memset(tmp1,0,sizeof(tmp1)); memset(tmp2,0,sizeof(tmp2)); tips：函数解释 void *memset(void *s, int ch, size_t n); 函数解释：将s中前n个字节替换为ch并返回s； cin>>tmp1>>tmp2; 量 数 组 长 度： len1 = strlen(tmp1); len2 = strlen(tmp2); for(i=len1-1,k=0;i>=0;i--)||改类型并换顺序 add1[k++] = tmp1[i] - '0'; for(i=len2-1,k=0;i>=0;i--) add2[k++] = tmp2[i] - '0'; tmp = 0; if(len1 >= len2) { for(i=0;i<=len1;i++) { res[i] = (add1[i] + add2[i] +tmp)%10; tmp = (add1[i] + add2[i] +tmp)/10; } } else if(len1 < len2) { for(i=0;i<=len2;i++) { 进位处理方法： res[i] = (add1[i] + add2[i] +tmp)%10; tmp = (add1[i] + add2[i] +tmp)/10; } } if(len1 >= len2) len = len1; else len = len2; printf("Case %d: %s + %s = ",j+1, tmp1 , tmp2); if(res[len]!=0) printf("%d",res[len]); for(i=len-1;i>=0;i--) printf("%d",res[i]); printf(" "); if(j!=N-1) printf(" "); } return 0; }
char tmp1[MAX], tmp2[MAX];
cin>>tmp1>>tmp2;
一长串的字符数组，还可以直接用cin进行输入
2. getchar（）必须加
3.字符型与整数类型互换
for(i=len1-1,k=0;i>=0;i--)
add1[k++] = tmp1[i] - '0';
for(i=len2-1,k=0;i>=0;i--)
add2[k++] = tmp2[i] - '0';
其实呢-48也是一样的；
4.错误代码
#include<iostream> #include<string.h> #include<stdio.h> using namespace std; int main() { int t; long sa,sb; cin>>t; getchar(); for(int i=1;i<=t;i++) { char a[5000],b[5000]; for(int j=0;;j++) { char word=getchar(); if(word>='0'&&word<='9') a[j]=word; else { sa=j; break; } } for(int k=0;;k++) { char word=getchar(); if(word>='0'&&word<='9') b[k]=word; else if(word==' ') { sb=k; break; } } cout<<"Case "<<i<<":"<<endl; for(int m=0;m<sa;m++) { cout<<a[m]; } cout<<" + "; for(int n=0;n<sb;n++) { cout<<b[n]; } cout<<" = "; int c[5000]={};编译有警告 int sc,next,shu,shua,shub; if(sa>sb) sc=sa; else sc=sb; 出错数据举例，999 999，所给出结果数组的长度设置有问题，应该分情况讨论，有0单独分情况，并设置不显示 for(int o=sc-1;o>=0;o--,sa--,sb--) { if(sa>0) shua=a[sa-1]-48; else shua=if(sb>0) shub=b[sb-1]-48; else shub=0; shu=shua+shub; if(shu>=10) { c[o-1]=1; shu=shu-10; c[o]+=shu; 进位不全面，1 99999999999，数据出错，所以不适宜顺序相加机制，应该采用逆序相加，顺序输出的方式 } else } for(int p=0;p<sc;p++) cout<<c[p]; cout<<endl; if(i!=t) cout<<endl; } #include<iostream>