VJ基础练习题二
U - The sum problem
Given a sequence 1,2,3,…N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.
Input
Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.
Output
For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.
Sample Input
20 10
50 30
0 0
Sample Output
[1,4]
[10,10]
[4,8]
[6,9]
[9,11]
[30,30]
输入N，M，求1-n的序列中，和是M的子序列。
Time Limit Exceeded
#include<iostream>
#include<string>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
int n,m;
while(cin>>n>>m){
if(n==0&&m==0){
break;
}
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
double sum = (j-i+1)*(j+i)/2;
if(sum == m){
cout<<"["<<i<<","<<j<<"]\n";
break;
}
if(sum > m){
break;
}
}
}
cout<<endl;
}
return 0;
}
Memory Limit Exceeded
#include<iostream>
#include<string>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
int n,m;
while(cin>>n>>m){
if(n==0&&m==0){
break;
}
double sum[n+1];
sum[0]=0;
for(int i=1;i<=n&&i<m;i++){
sum[i]=sum[i-1]+i;
}
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
double s = sum[j]-sum[i-1];
if(s == m){
cout<<"["<<i<<","<<j<<"]\n";
break;
}
if(s > m){
break;
}
}
}
cout<<endl;
}
return 0;
}
Accepted
#include<iostream>
using namespace std;
int main()
{
int n,m;
while(cin>>n>>m){
if(n==0&&m==0){
break;
}
for(int len=pow(2*m,0.5);len>=1;len--){
int beg = (2*m/len+1-len)/2;
if((beg*2+len-1)*len==2*m){
printf("[%d,%d]\n",beg,beg+len-1);
}
}
cout<<endl;
}
return 0;
}