程序贴出来了 sets: month/x1..x100/:a; year(month):b; links(month,year):f; endsets @bin(x); @for(links(i,j):12*a+b=100000);!说明这是有用的; @for(month:x+y=10000);!说明这是有用的; k=@min(links:f); fx=@if(x#le#1600,0, @if(x#le#2100,((X-1600)*0.05)*12, @if(x#le#3600,((X-2100)*0.1+500*0.05)*12, @if(x#le#3600,((X-2100)*0.1+500*0.05)*12, @if(x#le#6600,((x-3600)*0.15+1500*0.1+500*0.05)*12, @if(X#le#21600,((X-6600)*0.20+3000*0.15+1500*0.1+500*0.05)*12, @IF(x#le#41600,((x-21600)*0.25+20000*0.20+3000*0.15+1500*0.1+500*0.05)*12, @if(x#le#61600,((x-41600)*0.30+20000*0.25+20000*0.20+3000*0.15+1500*0.1+500*0.05)*12, @if(x#le#81600,((x-61600)*0.35+20000*0.30+20000*0.25+20000*0.20+3000*0.15+1500*0.1+500*0.05)*12, @if(x#le#100000,((x-81600)*0.40+20000*0.35+20000*0.30+20000*0.25+20000*0.20+3000*0.15+1500*0.1+500*0.05)*12, ((x-100000)*0.45+18400*0.40+20000*0.35+20000*0.30+20000*0.25+20000*0.20+3000*0.15+1500*0.1+500*0.05)*12 ))))))))))+ @if(y#le#6000,y*0.05, @if(y#le#24000,y*0.1, @if(y#le#60000,y*0.15, @if(y#le#240000,y*0.2,y*0.25)))); end 如果x10写成10w的话,一运行就超出内存,就取了个x10,y本来是表示年终将的,无奈,不知道该怎么加了,请教高手。 题目不难只是第一小题,但是程序把我难住了
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[此贴子已经被作者于2008-5-30 20:27:50编辑过] |