题解

453 条题解

  • 0
    @ 2007-07-23 08:10:34

    如果谁连这道题目都要用作弊来解决,那只能说明一个问题————这个人是个超级大傻瓜!

  • 0
    @ 2007-07-17 15:42:15

    var a:array[1..10];

    i,j,k:integer;

    begin

    for i:=1 to 10 do read(a[i]);

    readln(j);

    j:=j+30;

    k:=0;

    for i:=1 to 10 do

    begin

    if a[i]

  • 0
    @ 2007-08-05 16:20:16

    var

    a:array[1..10] of integer;

    i,x,n:integer;

    begin

    for i:=1 to 10 do read(a[i]);

    readln(n);

    n:=n+30;

    x:=0;

    for i:=1 to 10 do

    begin

    if a[i]

  • 0
    @ 2007-07-15 11:18:22

    var

    p,n,i:integer;

    a:array[1..10]of integer;

    begin

    for i:=1 to 10 do

    read(a[i]);

    readln(n);

    n:=n+30;

    p:=0;

    for i:=1 to 10 do

    if a[i]

  • 0
    @ 2007-07-15 11:07:49

    var

    p,n,i:integer;

    a:array[1..10]of integer;

    begin

    for i:=1 to 10 do

    read(a[i]);

    readln(n);

    n:=n+30;

    p:=0;

    for i:=1 to 10 do

    if a[i]

  • 0
    @ 2007-07-14 21:49:33

    var

    ans,i,n:longint;

    a:array[1..11] of longint;

    begin

    ans:=0;

    for i:=1 to 10 do read(a[i]);

    readln(n);

    for i:=1 to 10 do if n+30>=a[i] then inc(ans);

    writeln(ans);

    end.

  • 0
    @ 2007-07-14 20:33:44

    program pingguo;

    var

    a:array[1..10] of byte;

    i,j,p,n:byte;

    begin

    for i:=1 to 10 do

    read(a[i]);

    readln(n);

    p:=0;

    for i:=1 to 10 do

    if n+30>=a[i] then p:=p+1;

    writeln(p);

    end.

  • 0
    @ 2007-07-11 16:03:52

    program P1102;

    var

    a:array[1..10] of real;

    b,i,j,h:integer;

    begin

    j:=1;

    for i:=1 to 10 do read(a[i]);

    readln(h);

    for i:=1 to 10 do

    if h+30>a[i] then inc(j);

    writeln(j);

    end.

  • 0
    @ 2007-07-04 13:20:05

    62%了...

  • 0
    @ 2007-06-28 17:18:59

    大家共同创造AC神话!!

  • 0
    @ 2009-06-28 21:18:28

    最大流

    建立源点s,汇点t,插入一个点a,再插入十个点i1,i2...i10,如果A[i]>=k+30,连边G[a,i],容量为1。用ford-folkerson算法做即可。

    编译通过...

    ├ 测试数据 01:答案正确... 97ms

    ├ 测试数据 02:答案正确... 132ms

    ├ 测试数据 03:答案正确... 144ms

    ├ 测试数据 04:答案正确... 65ms

    ├ 测试数据 05:答案正确... 90ms

    ├ 测试数据 06:答案正确... 31ms

    ├ 测试数据 07:答案正确... 17ms

    ├ 测试数据 08:答案正确... 273ms

    ├ 测试数据 09:答案正确... 84ms

    ├ 测试数据 10:答案正确... 46ms

    ---|---|---|---|---|---|---|---|-

    Accepted 有效得分:100 有效耗时:979ms

    效率不是很高,但也能过...

    [请神牛bs我]

  • 0
    @ 2007-06-22 21:13:47

    var

    ans,i,n:longint;

    a:array[1..11] of longint;

    begin

    ans:=0;

    for i:=1 to 10 do read(a[i]);

    readln(n);

    for i:=1 to 10 do if n+30>=a[i] then inc(ans);

    writeln(ans);

    end.

  • 0
    @ 2007-06-11 19:02:02

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 0ms

    ├ 测试数据 05:答案正确... 0ms

    ├ 测试数据 06:答案正确... 0ms

    ├ 测试数据 07:答案正确... 0ms

    ├ 测试数据 08:答案正确... 0ms

    ├ 测试数据 09:答案正确... 0ms

    ├ 测试数据 10:答案正确... 0ms

    ---|---|---|---|---|---|---|---|-

    Accepted 有效得分:100 有效耗时:0ms

  • 0
    @ 2007-06-03 09:42:11

    var

    a:array[1..11]of integer;

    t,s:integer;

    begin

    for t:=1 to 11 do read(a[t]);

    begin

    for t:=1 to 10 do

    if a[t]

  • 0
    @ 2007-06-08 16:03:15

    var

    a:array[1..12]of integer;

    m,s,x:integer;

    begin

    for s:=1 to 12 do read(a);

    read(m);

    m:=m+30;

    for s:=1 to 12 do

    if a

  • 0
    @ 2007-05-27 12:18:19

    就两字,弱智!

  • 0
    @ 2007-05-20 16:41:25

    program aaa;

    var

    a:array[1..10]of integer;

    n,i,t:integer;

    begin

    for i:=1 to 10 do

    read(a[i]);

    readln(n); t:=0;

    for i:=1 to 10 do

    if n+30>=a[i]

    then t:=t+1;

    writeln(t);

    end.

  • 0
    @ 2007-04-26 16:40:47

    初三比赛的题,拿到高一做,17行就解决了。。。是简洁了好多。。

  • 0
    @ 2007-04-25 22:24:49

    郁闷 搞什么..............对了 说什么无评测结果 气死我

  • 0
    @ 2007-03-06 13:31:07

    var

    a,b,i,j:longint;

    begin

    readln(a);

    readln(b);

    i:=0;

    for i:=1 to 10 do

    if a

信息

ID
1102
难度
3
分类
模拟 点击显示
标签
递交数
16729
已通过
8792
通过率
53%
被复制
58
上传者