抓起根本(二)(hdu 4554 叛逆的小明 hdu 1002 A + B Problem II,数字的转化(反转),大数的加法......)
2015-03-02 13:41
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数字的反转:
就是将数字倒着存下来而已。(*^__^*) 嘻嘻……
大致思路:将数字一位一位取出来,存在一个数组里面,然后再将其变成数字,输出。
详见代码。
趁热打铁 例题:hdu 4554 叛逆的小明
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4554
Total Submission(s): 818 Accepted Submission(s):
568
[align=left]Problem Description[/align]
叛逆期的小明什么都喜欢反着做,连看数字也是如此(负号除外),比如:
小明会把1234它看成4321;把-1234看成-4321;把230看成032
(032=32);把-230看成-032(-032=-32)。
现在,小明做了一些a+b和a-b的题目(a,
b为整数且不含前导0),如果给你这些题目正确的答案,你能猜出小明会做得到什么答案吗?
[align=left]Input[/align]
输入第一行为一个正整数T(T<=10000),表示小明共做了T道题。
接下来T行,每行是两个整数x,y(-1000000<=x,
y<=1000000), x表示a+b的正确答案,y表示a-b的正确答案。
输入保证合法,且不需考虑a或b是小数的情况。
[align=left]Output[/align]
输出共T行,每行输出两个整数s
t,之间用一个空格分开,其中s表示小明将得到的a+b答案,t表示小明将得到的a-b答案。
[align=left]Sample Input[/align]
3
20 6
7 7
-100 -140
[align=left]Sample Output[/align]
38 24
7 7
-19 -23
题目大意:中文题目,清晰明了~
详见代码。
大数加法:
数字很大,假如直接加。很容易超时!所以换一种方法就是直接取出来一位一位加,不过不能换成数字。
例题:hdu 1002 A + B Problem II
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002
Total Submission(s): 238717 Accepted Submission(s):
46010
[align=left]Problem Description[/align]
I have a very simple problem for you. Given two
integers A and B, your job is to calculate the Sum of A + B.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
2
1 2
112233445566778899 998877665544332211
[align=left]Sample Output[/align]
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
参考以上的详解。将数字一位一位取出来,倒过来相加,在输出~
就是将数字倒着存下来而已。(*^__^*) 嘻嘻……
大致思路:将数字一位一位取出来,存在一个数组里面,然后再将其变成数字,输出。
详见代码。
while (a) //将每位数字取出来,取完为止 { num1[i]=a%10; //将每一个各位取出存在数组里面,实现了将数字反转 i++; //数组的变化 a/=10; }
趁热打铁 例题:hdu 4554 叛逆的小明
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4554
叛逆的小明
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 818 Accepted Submission(s):
568
[align=left]Problem Description[/align]
叛逆期的小明什么都喜欢反着做,连看数字也是如此(负号除外),比如:
小明会把1234它看成4321;把-1234看成-4321;把230看成032
(032=32);把-230看成-032(-032=-32)。
现在,小明做了一些a+b和a-b的题目(a,
b为整数且不含前导0),如果给你这些题目正确的答案,你能猜出小明会做得到什么答案吗?
[align=left]Input[/align]
输入第一行为一个正整数T(T<=10000),表示小明共做了T道题。
接下来T行,每行是两个整数x,y(-1000000<=x,
y<=1000000), x表示a+b的正确答案,y表示a-b的正确答案。
输入保证合法,且不需考虑a或b是小数的情况。
[align=left]Output[/align]
输出共T行,每行输出两个整数s
t,之间用一个空格分开,其中s表示小明将得到的a+b答案,t表示小明将得到的a-b答案。
[align=left]Sample Input[/align]
3
20 6
7 7
-100 -140
[align=left]Sample Output[/align]
38 24
7 7
-19 -23
题目大意:中文题目,清晰明了~
详见代码。
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int num1[10],num2[10]; int main () { int T; scanf("%d",&T); while (T--) { int x,y,a,b,i=0,j=0; int p=0,q=0; scanf("%d%d",&x,&y); memset(num1,0,sizeof(num1)); memset(num2,0,sizeof(num2)); a=(x+y)/2; b=x-a; while (a) { num1[i]=a%10; i++; a/=10; } while (b) { num2[j]=b%10; j++; b/=10; } for (int k=0; k<i; k++) //换成数字 p=num1[k]+p*10; for (int l=0; l<j; l++) q=num2[l]+q*10; printf ("%d %d\n",p+q,p-q); } return 0; }
大数加法:
数字很大,假如直接加。很容易超时!所以换一种方法就是直接取出来一位一位加,不过不能换成数字。
int l=l1>l2?l1:l2; //在两个之间取一个最长的 for (int i=0; i<l; i++) { num3[i]=num1[i]+num2[i]; //一位一位加 if (num3[i-1]>9&&i>=1) //考虑进位的问题,如果大于9就需要进位 { num3[i]++; } } if (num[l-1]>9) //最后一位的进位问题 { num[l]++; //仍需要进位 l++; //长度需要加一 }
例题:hdu 1002 A + B Problem II
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 238717 Accepted Submission(s):
46010
[align=left]Problem Description[/align]
I have a very simple problem for you. Given two
integers A and B, your job is to calculate the Sum of A + B.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
2
1 2
112233445566778899 998877665544332211
[align=left]Sample Output[/align]
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
参考以上的详解。将数字一位一位取出来,倒过来相加,在输出~
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main () { int n,flag=1; char a[8000],b[8000]; int num1[8000],num2[8000],ans[8000]; scanf("%d",&n); while (n--) { int k=0,s=0; scanf("%s%s",a,b); printf ("Case %d:\n",flag++); int len1=strlen(a); int len2=strlen(b); memset(num1,0,sizeof(num1)); memset(num2,0,sizeof(num2)); memset(ans,0,sizeof(ans)); int l=len1>len2?len1:len2; for (int i=len1-1; i>=0; i--) { num1[k]=a[i]-'0'; k++; } for (int j=len2-1; j>=0; j--) { num2[s]=b[j]-'0'; s++; } for (int i=0; i<l; i++) { ans[i]=num1[i]+num2[i]; if (i>=1) if (ans[i-1]>9) { ans[i]++; } //cout<<num1[i]<<" "<<num2[i]<<" "<<ans[i]<<endl; } if (ans[l-1]>9) { ans[l]=1; l++; } printf ("%s + %s = ",a,b); for (int i=l-1; i>=0; i--) printf("%d",ans[i]%10); if (n) printf ("\n"); printf ("\n"); } return 0; }
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