思路是计算两个扇形的面积和 $sum1$，减去筝形的面积 $sum2$。

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<cstdlib>
#include<ctime>
#define rep(i,a,b) for(register int i=a;i<=b;++i)
#define rev(i,a,b) for(register int i=a;i>=b;--i)
#define gra(i,u) for(register int i=head[u];i;i=edge[i].nxt)
#define Clear(a) memset(a,0,sizeof(a))
#define yes puts("YES")
#define no puts("NO")
using namespace std;
typedef long double ll;
#define y1 ptsd
#define y2 zyz
const int INF(1e9+10);
const ll LLINF(1e18+10);
inline int read()
{
    int s=0,w=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')s=s*10+(ch-'0'),ch=getchar();
    return s*w;
}
template<typename T>
inline T Min(T x,T y){return x<y?x:y;}
template<typename T>
inline T Max(T x,T y){return x>y?x:y;}
template<typename T>
inline void Swap(T&x,T&y){T t=x;x=y;y=t;return;}
template<typename T>
inline T Abs(T x){return x<0?-x:x;}

ll x1,y1,r1;
ll x2,y2,r2;

inline ll D(){return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);}
const ll pi(3.141592654);
inline ll sq(ll x){return x*x;}

int main()
{
    // freopen("read.txt","r",stdin);
    x1=read(),y1=read(),r1=read();
    x2=read(),y2=read(),r2=read();
    ll d=sqrt(D());
    if(r1+r2<d) puts("0");
    else if(Abs(r1-r2)>=d) printf("%.6Lf\n",Min(r1,r2)*Min(r1,r2)*pi);
    else
    {
    	puts("YES");
    	ll c1x=(sq(r1)+sq(d)-sq(r2))/(2*r1*d);
    	ll c2x=(sq(r2)+sq(d)-sq(r1))/(2*r2*d);
    	c1x=acos(c1x)*2.0,c2x=acos(c2x)*2.0;
    	ll a=c1x*sq(r1)/2,b=c2x*sq(r2)/2;
    	ll sum1=a+b;
    	ll h=sq(r2)+sq(r2)-2*r2*r2*cos(c2x);
    	h=sqrt(h);
    	ll sum2=h*d/2;
    	printf("%.6Lf\n",sum1-sum2);

    }
    return 0;
}