hdoj 1002 A + B Problem II 【大數A+B】
阿新 • • 發佈:2019-01-30
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 308540 Accepted Submission(s): 59647
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.
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題解:大數相加,將兩個字串反轉 (s1[i]-'0' s2[i]-'0'),存於a[] b[],然後逐位相加,判斷進位,倒序輸出!!!注意格式!!!
#include <cstdio> #include <cstring> #include <cmath> char s1[1010],s2[1010]; int a[1010],b[1010],c[1010]; int max(int a,int b) { if(a>b) return a; else return b; } int main() { int t,k=1; scanf("%d",&t); while(t--) { scanf("%s%s",s1,s2); printf("Case %d:\n",k++); printf("%s + %s = ",s1,s2); memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); int la=strlen(s1); int lb=strlen(s2); int ans=0,cnt=0; for(int i=la-1;i>=0;i--) { a[ans]=s1[i]-'0';//字串反轉 ans++; } for(int i=lb-1;i>=0;i--) { b[cnt]=s2[i]-'0'; cnt++; } int maxn; maxn=max(la,lb); for(int i=0;i<=maxn-1;i++)//從左到右逐位相加 { a[i]=a[i]+b[i];//將各位之和還賦給陣列a[] if(a[i]>=10)//考慮各位之和大於等於10的情況,考慮進位 { a[i]-=10; a[i+1]++; } } if(a[maxn]==0)//判斷最後一位是否有進位 { for(int i=maxn-1;i>=0;i--) printf("%d",a[i]); } else { for(int i=maxn;i>=0;i--) printf("%d",a[i]); } if(t==0) printf("\n"); else printf("\n\n");//每行輸出間有空格 } return 0; }
再附上一個屬於我的!用c[] 儲存結果
#include <cstdio> #include <cmath> #include <cstring> char s1[1010],s2[1010]; int a[1010],b[1010],c[1010]; int max(int a,int b) { if(a>b) return a; else return b; } int main() { int t,k=0; scanf("%d",&t); while(t--) { scanf("%s%s",s1,s2); printf("Case %d:\n",k++); printf("%s + %s = ",s1,s2); memset(a,0,sizeof(a));//清零!!! memset(b,0,sizeof(b));//清零 !!! memset(c,0,sizeof(c));//清零!!! int la,lb; la=strlen(s1); lb=strlen(s2); int ans=0,cnt=0; for(int i=la-1;i>=0;i--) { a[ans]=s1[i]-'0'; ans++; } for(int i=lb-1;i>=0;i--) { b[cnt]=s2[i]-'0'; cnt++; } int maxn=max(la,lb); for(int i=0;i<=maxn-1;i++) { if((c[i]+a[i]+b[i])>=10) c[i+1]++; c[i]=(c[i]+a[i]+b[i])%10; } if(c[maxn]==0) { for(int i=maxn-1;i>=0;i--) printf("%d",c[i]); } else { for(int i=maxn;i>=0;i--) printf("%d",c[i]); } if(t==0) printf("\n"); else printf("\n\n"); } return 0; }