hdu 3064

1：前n项和公式：1+2+3+...+n = n*(n+1)/2

2：前n项平方和公式：1^2+2^2+.........+n^2=n*（n＋1）*（2n＋1）/6

#include<stdio.h>

#include<math.h>

int main() {

__int64 t,x,y;

__int64 ans,sum;

double a,b;

while(scanf("%I64d",&t)!=EOF) {

scanf("%I64d%I64d",&x,&y);

sum=(x+y)*(y-x+1)/2;

ans=y*(y+1)*(2*y+1)/6-(x-1)*x*(x*2-1)/6;

while(t--) {

scanf("%d%d",&x,&y);

ans=ans-y*(y+1)*(2*y+1)/6+(x-1)*x*(x*2-1)/6;

sum=sum-(x+y)*(y-x+1)/2;

}

a=(sum+sqrt(1.0*sum*sum-4.0*(sum*sum-ans)/2))/2;

b=(sum-sqrt(1.0*sum*sum-4.0*(sum*sum-ans)/2))/2;

 printf("%.0f %.0f\n",b,a);

}

return 0;

}