Input
杈撳叆鐨勬瘡涓琛屾湁鍗曠嫭鐨勪竴涓糿銆俷=0鏍囧織杈撳叆緇撴潫Output
瀵逛簬杈撳叆鐨勬瘡涓琛岋紝浣跨敤鍗曠嫭涓琛岃緭鍑哄搴旂粨鏋溿?Sample Input
2 5 10 0
Sample Output
1 3 8鎬葷粨瑙勫緥錛?br>1錛?
2錛?
3錛?
4錛?
5錛?
6錛?
7錛?
8錛?
9錛?
10錛?
11錛?
12錛?0
16錛?5
32錛?1
64錛?3
浣犱細鍙戠幇2^n鐨勯兘鏄?^n-1
鍏朵粬13=8+4+1=(8-1)+(4-1)+(1-1),鏈夊灝戜釜2^n,鍒欏噺澶氬皯嬈?銆?br>鐩存帴鏍規嵁浣嶈繍綆楀嵆鍙眰
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For Example, when N = 2 and M = 3, there are 6 squares size of 1 * 1 and 2 squares size of 2 * 2 in the chessboard, so the answer is 8.
Input
There are several lines in the input, each line consist of two positive integers N and M (1 <= N <= 100, 1 <= M <= 100) except the last line which N = 0 and M = 0 implying the end of the input and you should not process it.Output
Please output the number of the squares for each chessboard in a single line.Sample Input
1 1 2 3 3 3 0 0
Sample Output
1 8 14
鍚彂錛氫漢鏄庝箞綆楃殑錛屽氨璁╃▼搴忔庝箞綆楋紝涓姝ヤ竴姝ユ帹銆?br>#include<iostream>
using namespace std;
int main()
{
int n,m;
long sum=0;
int i;
while(cin>>n>>m,n)
{
if(n==m)
{
sum=n*(n+1)*(2*n+1)/6;
cout<<sum<<endl;
}
else
{
for(i=0;i<n&&i<m;i++)
sum+=(n-i)*(m-i);
cout<<sum<<endl;
}
sum=0;
}
return 0;
}
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In arithmetic we often use the scheme called 'round up by five' to round up a decimal fraction number into a decimal integer number whose fraction part equals to or is bigger than 0.5, otherwise, its fraction part is discarded.
In this problem, you are to implement a similar sheme called 'round up by six'.
Input
The first line of the input is an integer N indicating the number of cases you are to process. Then from the second line there are N lines, each containing a number (may or may not be fractional).
Output
For each number in the input, you are to print the round-up-by-six version of the number in a single line.
Sample Input
2 25.599 0.6
Sample Output
25 1
#include<iostream>
#include<cmath>
using namespace std;
int main()
{int n;
cin>>n;
while(n--)
{ double a;
cin>>a;
double b=floor(a);
if(a-b>=0.6)
cout<<b+1<<endl;
else
cout<<b<<endl;
}
return 0;
}
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As you known,a positive integer can be written to a form of products of several positive integers(N = x1 * x2 * x3 *.....* xm (xi!=1 (1<=i<=m))).Now, given a positive integer, please tell me how many forms of products.
Input
Input consists of several test cases.Given a positive integer N (2<=N<=200000) in every case.
Output
For each case, you should output how many forms of the products.
Sample Input
12 100 5968
Sample Output
4 9 12
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
int a[200001];
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int i,j,n;
memset(a,0,sizeof(a));
a[1]=1;
for(i=1;i<=200000;i++)
for(j=1;i*j<=200000;j++)
a[i*j]+=a[j];
while(scanf("%d",&n)!=EOF)
printf("%d\n",a[n]/2);
//system("PAUSE");
return 0;
}
鍙戠幇閫掓帹瑙勫緥瑕佷粠鐗圭偣鍜屽嚭鍙戯紝涔樻硶鑷劧鏄箻娉曘?
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