学步园

TAG 数论

与1209. Sequence Sum Possibi 类似，我是用那道题改的。

#include <stdio.h><br /> #include <math.h><br /> #include <algorithm><br /> using namespace std;<br /> struct node<br /> {<br /> int a,b;<br /> };<br /> int T;<br /> int n,m;<br /> int counter;<br /> node ans[1000];<br /> bool cmp( const node &x, const node &y)<br /> {<br /> if ( x.a!=y.a )<br /> {<br /> return x.a<y.a;<br /> }<br /> return x.b<y.b;<br /> }<br /> void cal()<br /> {<br /> counter=-1;<br /> for (int i=0; i<=int( sqrt(2.0)*sqrt(m+0.0) ); ++i)<br /> {<br /> if ( (2*m)%(i+1)==0 && (2*m/(i+1)-i) >=2 && (2*m/(i+1)-i)%2==0 )<br /> {<br /> int start=2*m/(i+1)-i;<br /> start/=2;<br /> if ( i+start<=n )<br /> {<br /> ++counter;<br /> ans[ counter ].a=start;<br /> ans[ counter ].b=start+i;<br /> }<br /> }<br /> }<br /> }<br /> int main(int argc, char *argv[])<br /> {<br /> scanf("%d%d", &n, &m);<br /> cal();<br /> sort( ans, ans+counter+1,cmp );<br /> for (int i=0; i<=counter; ++i)<br /> {<br /> printf("[%d,%d]/n", ans[i].a, ans[i].b );<br /> }<br /> return 0;<br /> }<br />