求PASCAL 语言的解题步骤,
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求PASCAL 语言的解题步骤,
var
a,b:array[0..2] of integer;
i,j,tmp:integer;
begin
for i := 0 to 2 do
read(b[i]);
for i := 0 to 2 do
begin
a[i] := 0;
for j := 0 to i do
begin
inc(a[i],b[j]);
inc(b[a[i] mod 3],a[j]);
end;
end;
tmp := 1;
for i := 0 to 2 do
begin
a[i] := a[i] mod 10;
b[i] := b[i] mod 10;
tmp := tmp * (a[i] + b[i]);
end;
writeln(tmp);
end.
输入:2 3 5
输出:_______
var
a,b:array[0..2] of integer;
i,j,tmp:integer;
begin
for i := 0 to 2 do
read(b[i]);
for i := 0 to 2 do
begin
a[i] := 0;
for j := 0 to i do
begin
inc(a[i],b[j]);
inc(b[a[i] mod 3],a[j]);
end;
end;
tmp := 1;
for i := 0 to 2 do
begin
a[i] := a[i] mod 10;
b[i] := b[i] mod 10;
tmp := tmp * (a[i] + b[i]);
end;
writeln(tmp);
end.
输入:2 3 5
输出:_______
![求PASCAL 语言的解题步骤,](/uploads/image/z/18510643-19-3.jpg?t=%E6%B1%82PASCAL+%E8%AF%AD%E8%A8%80%E7%9A%84%E8%A7%A3%E9%A2%98%E6%AD%A5%E9%AA%A4%2C)
这是NOIP普及组第15届的一道题目,答案是416,需要注意的两点是a,b数组都是从0开始的,inc(b[a[i] mod 3], a[j]); 就是b[a[i] mod 3]:=b[a[i] mod 3]+a[j].其他的自己列两个表分别表示两个数组跟踪就行了,没什么难度.