[1303] A + B
時間限制: 1000 ms 內存限制: 65535 K
問題描述
As we all know, in the computer science, an integer A is in the range of 32-signed integer, which means the integer A is between -2^31 and (2^31)-1 (inclusive) and A is a 64-signed integer, which means A is between -2^63 and (2^63)-1(inclusive). Now we give the K-signed range, and two K-signed integers A and B, you should check whether the sum of A and B is beyond the range of K-signed integer or not.
輸入
There will be many cases to calculate. In each case, there comes the integer K (2<=K<=64) first in a single line.
Then following the line, there is another single line which has two K-signed integers A and B.
輸出
For each case, you should estimate whether the sum is beyond the range. If exceeded, print “Yes”, otherwise “I love nbut”.
樣例輸入
64
1000 1000樣例輸出
I love nbut提示
無來源
加多寶涼茶操作
http://acm.nbut.cn/Problem/view.xhtml?id=1303
題意 ：輸入k a b 表示 k位大小的數據 如 int 是32位 long long 是 64位 k最大為64 輸入a b 問a+b的和是否溢出
思路：
我們用long long 存儲a b 對於k<64 a+b不會溢出 直接根據k位數據范圍判斷
當 k為64 則可能溢出 我們可以用 a+b=c 判斷 c-a是不是等於b c-b是不是等於a
或者用最下面的方法看代碼 2種處理都不錯
[cpp]
#include<stdio.h>
#include<math.h>
int main()
{
int k;
__int64 a,b,c,left,right,max,min;
while(scanf("%d",&k)!=EOF)
{
scanf("%I64d %I64d",&a,&b);
if(k==64)
{
c=a+b;
if(c-a!=b||c-b!=a)
{
printf("Yes\n");
}
else printf("I love nbut\n");
continue;
}
left=-1;right=1;
int i;
for(i=1;i<=k-1;i++)
{
left*=2;
right*=2;
}
right-=1;
c=a+b;
if(left<=c&&c<=right)
{
printf("I love nbut\n");
}
else printf("Yes\n");
}
return 0;
}
#include<stdio.h>
#include<math.h>
int main()
{
int k;
__int64 a,b,c,left,right,max,min;
while(scanf("%d",&k)!=EOF)
{
scanf("%I64d %I64d",&a,&b);
if(k==64)
{
c=a+b;
if(c-a!=b||c-b!=a)
{
printf("Yes\n");
}
else printf("I love nbut\n");
continue;
}
left=-1;right=1;
int i;
for(i=1;i<=k-1;i++)
{
left*=2;
right*=2;
}
right-=1;
c=a+b;
if(left<=c&&c<=right)
{
printf("I love nbut\n");
}
else printf("Yes\n");
}
return 0;
}
[cpp]
#include<stdio.h>
#include<math.h>
int main()
{
int k;
__int64 a,b,c,left,right,max,min;
while(scanf("%d",&k)!=EOF)
{
scanf("%I64d %I64d",&a,&b);
if(k==64)
{
max=(__int64)pow(2.0,k-1.0)-1;
min=(__int64)pow(2.0,k-1.0);
if(a>=0&&b<=0||a<=0&&b>=0)
printf("I love nbut\n");
else
{
if(a>0&&b>0)
{
a=a-max+b;
if(a>0)
printf("Yes\n");
else
printf("I love nbut\n");
}
else if(a<0&&b<0)
{
a=a-min+b;
if(a<0)
printf("Yes\n");
else
printf("I love nbut\n");
}
}
continue;
}
left=-1;right=1;
int i;
for(i=1;i<=k-1;i++)
{
left*=2;
right*=2;
}
right-=1;
c=a+b;
if(left<=c&&c<=right)
{
printf("I love nbut\n");
}
else printf("Yes\n");
}
return 0;
}
#include<stdio.h>
#include<math.h>
int main()
{
int k;
__int64 a,b,c,left,right,max,min;
while(scanf("%d",&k)!=EOF)
{
scanf("%I64d %I64d",&a,&b);
if(k==64)
{
max=(__int64)pow(2.0,k-1.0)-1;
min=(__int64)pow(2.0,k-1.0);
if(a>=0&&b<=0||a<=0&&b>=0)
printf("I love nbut\n");
else
{
if(a>0&&b>0)
{
a=a-max+b;
if(a>0)
printf("Yes\n");
else
printf("I love nbut\n");
}
else if(a<0&&b<0)
{
a=a-min+b;
if(a<0)
printf("Yes\n");
else
printf("I love nbut\n");
}
}
continue;
}
left=-1;right=1;
int i;
for(i=1;i<=k-1;i++)
{
left*=2;
right*=2;
}
right-=1;
c=a+b;
if(left<=c&&c<=right)
{
printf("I love nbut\n");
}
else printf("Yes\n");
}
return 0;
}
