用了等差数列求和，呵呵

#include<stdio.h>
int main(){
int M;
scanf("%d",&M);
for (int i=1;i<=M/2;i++){
for (int j=i+1;(i+j)*(j-i+1)<=2*M;j++){
if ((i+j)*(j-i+1)==2*M) { //（首项+末项）*项数/2
printf("%d %d\n",i,j);
break;
}
}
}
return 0;
}