设f(x)=ax^2+bx+cf(0)=c=1f(x+1)=a(x+1)^2+b（x+1）+c=ax^2+(2a+b)x+a+c+bf(x)+x+1=ax^2+(b+1)x+c+1ax^2+(2a+b)x+a+c+b=ax^2+(b+1)x+c+1所以2a+b=b+1,a+c+b=c+1,而c=1则a=1/2,b=1/2,c=1f(x)=1/2x^2+1/2x+1