B - Sum Problem
In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.
Input
The input will consist of a series of integers n, one integer per line.
Output
For each case, output SUM(n) in one line, followed by a blank line.
You may assume the result will be in the range of 32-bit signed integer.
Sample Input
1
100
Sample Output
1
5050
WrongAnswer
#include <stdio.h>
int main() {
int n,sum;
while(scanf("%d",&n)!=EOF){
*sum=((1+n)*n)/2;* //因为(1+n)*n时，当数据过大时会造成数据溢出，从而出现WA。
printf("%d
",sum);
}
return 0;
}
修改后：
#include<stdio.h>
int main(){
int n,sum1;
while(scanf("%d",&n)!=EOF){
*sum1=n/2*(n+1);*
printf("%d
",sum1);
}
return 0;
}
C - A + B Problem II
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
WrongAnswer
这题我首先就注意到了第二组数据过大，所以我写了一个数据类型为long long的代码，如下：
#include<stdio.h>
int main()
{
int n,i;
long long a=0,b =0;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%lld%lld",&a,&b);
printf("Case %d:
%lld + %lld = %lld",i,a,b,a+b);
if(i!=n){printf("
");}//这里要注意格式！！
}
return 0;
}
但是很快发现时WA，后面我意识到需要用字符串进行解题，也就是大数问题（使2字符串的中的字符数字减去'0'，逐个相加大于等于10的可以使本位减10，下一位自增1，后面的处理就非常简单了；）
#include<stdio.h>
#include<string.h>
int num(int sum[],int len1,int len2,int j,char str1[],char str2[]);
int main()
{
int n,i,j;
char str1[1010],str2[1010];
long t;
long len1,len2;
scanf("%d",&n);
for(i=1;i<=n;i++){
int flag=0;
int sum[10000]={0};
scanf("%s %s",str1,str2);
printf("Case %d:
%s + %s = ",i,str1,str2);
len1=strlen(str1)-1;
len2=strlen(str2)-1;
j=0;
while(len1>=0&&len2>=0){
if(sum[j]+(str1[len1]-'0')+(str2[len2]-'0')>=10){//逢十进一
sum[j]=sum[j]+(str1[len1]-'0')+(str2[len2]-'0')-10;
sum[j+1]++;
}
else{
sum[j]=sum[j]+(str1[len1]-'0')+(str2[len2]-'0');
}
j++;len1--;len2--;}
if(len1>=0){
for(t=len1;t>=0;t--){
sum[j]=sum[j]+(str1[t]-'0');
j++;
}
}
else if(len2>=0){
for(t=len2;t>=0;t--){
sum[j]=sum[j]+(str2[t]-'0');
j++;
}
}
else if(sum[j]!=0) j++;//两个位数相同的数，这步不能丢！
for(t=j-1;t>=0;t--){
if(sum[t]==0&&flag==0) continue;
else{
flag=1;
printf("%d",sum[t]);
}
}
if(i!=n){printf("
");}//注意输出格式
else printf("
");
}
return 0;
}
[参考]--问题 C: A+B Problem II