请问如何简化以下code中的do循环累加语句（红色的code）？谢谢！

说明：该段code运行结果正确.


data gcb2;
    set gcb1;

    array food(9) food1-food9;
    do i = 1 to 9;
        *kindexc()是用于识别单个汉字的，
         而indexc()是用于识别单个字节的（当用indexc()识别汉字时会生成错误的分类结果但log不会报错）;
        if kindexc(food(i),"菜","瓜","竹","柿","葱","蒜","卜","椒","冬","姜")  then food(i)="蔬菜";
        else if kindexc(food(i),"肉","骨","脚","腿","翅","鱼","尖")  then food(i)="禽畜肉";
        else if kindexc(food(i),"面","米","粉","包") then food(i)="粮谷类";
        else if kindexc(food(i),"蛋")  then food(i)="鸡蛋";
        else if kindexc(food(i),"奶")  then food(i)="牛奶";
        else if kindexc(food(i),"油")  then food(i)="植物油";
        else if indexc(food(i),"盐")  then food(i)="盐";
        else if indexc(food(i),"抽") then food(i)="生抽";
    end;

   array effective(9) effective1-effective9;
   do i = 1 to 9;
   do j = 1 to 9;
       if  kindexc(food(i),"黄","大")  then  effective(j)=effective(j)*4;
   end;
   end;

   do i = 1 to 9;
    if kindexc(food(i),"黄","大","水","豆")  then food(i)="豆腐";
   end;

   *ef= expected food,ef1-ef7依次为粮谷类、蔬菜、禽畜肉、鸡蛋、豆腐、牛奶、植物油,单位为“斤”;
           ef1 = (n1*330 + n2*375 + n3*475)/500;
        ef2 = (n1*450 + n2*475 + n3*500)/500;
         ef3 = (n1*60 + n2*75 + n3*125)/500;
        ef4 = (n1*1 + n2*1.5 + n3*1.5)*60/500;
        ef5 = (n1*150 + n2*200 + n3*200)/500;
         ef6 = (n1*150 + n2*150 + n3*150)/500;
        ef7 = (n1*25 + n2*25 + n3*25)/500;
   
    *af= actual food,af1-af9依次为粮谷类、蔬菜、禽畜肉、鸡蛋、豆腐、牛奶、植物油、盐、生抽,单位为“斤”;
    do i = 1 to 9;
        if food(i)="粮谷类" then do;
            af1=0;
            af1+effective(i);
     end;
    end;

    do i = 1 to 9;
        if food(i)="蔬菜" then do;
            af2=0;
            af2+effective(i);
    end;
    end;

    do i = 1 to 9;
        if food(i)="禽畜肉" then do;
            af3=0;
            af3+effective(i);
    end;
    end;

    do i = 1 to 9;
        if food(i)="鸡蛋" then do;
            af4=0;
            af4+effective(i);
     end;
    end;

    do i = 1 to 9;
        if food(i)="豆腐" then do;
            af5=0;
            af5+effective(i);
    end;
    end;

    do i = 1 to 9;
        if food(i)="牛奶" then do;
            af6=0;
            af6+effective(i);
    end;
    end;


    do i = 1 to 9;
        if food(i)="植物油" then do;
            af7=0;
            af7+effective(i);
     end;
    end;

    do i = 1 to 9;
        if food(i)="盐" then do;
            af8=0;
            af8+effective(i);
    end;
    end;

    do i = 1 to 9;
        if food(i)="生抽" then do;
            af9=0;
            af9+effective(i);
    end;
    end;

    run;