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题解

28 条题解

  • 2
    @ 2017-10-18 21:24:24
    //比较简单的模拟题,考察的是取模的应用
    #include<iostream>
    using namespace std;
    int room[10002][102],pan[10002][102],lou[10002][102];
    int main()
    {
        int n,m,i,k,ans=0,now,bian,sum=0;
        cin>>n>>m;
        for(i=1;i<=n;i++)
         {
            int has=0;
         for(k=0;k<m;k++)
          {
           cin>>pan[i][k]>>room[i][k];
           if(pan[i][k])
            {
                has++;
                lou[i][101]++;
                lou[i][has]=k;
            }
          }
         }
        cin>>now;
        for(i=1;i<=n;i++)
        {
            sum+=room[i][now];
            sum%=20123;
            for(k=1;k<=lou[i][101];k++)
             if(lou[i][k]>=now)
              break;
             if(k>lou[i][101])
              k=1;
            bian=(room[i][now]-1)%lou[i][101];
            now=lou[i][(k+bian-1)%lou[i][101]+1];   
        }
        cout<<sum;
        return 0;
    }
    
  • 1
    @ 2021-08-30 06:42:07
    #include<bits/stdc++.h>
    using namespace std;
    
    int a[10010][110],bk[10010][110],dd[10010][110],quan[10010]; 
    int lm,n,m,ff,sum,nn=0;
    int main(){
        cin>>n>>m;
        memset(a,0,sizeof(a));
        for(int i=1; i<=n; i++)
            for(int j=0; j<m; j++){
                cin>>bk[i][j]>>a[i][j];
                quan[i]=quan[i]+bk[i][j];            
            }
        cin>>ff;
        for (int i=1; i<=n; i++){
            sum=sum+a[i][ff];
            lm=a[i][ff];
            lm=lm%quan[i]+quan[i];
            lm=lm-bk[i][ff];
            while (lm>0){
                ff++;
                ff=ff%m;
                lm=lm-bk[i][ff];
            }
        }    
        cout<<sum%20123;
        return 0;
    }
    
  • 0
    @ 2021-02-25 12:27:33

    #include<bits/stdc++.h>
    using namespace std;
    int n,m,k,t,a[10002][102],b[10002][102],l,c[10002],d;
    long long int ans;
    int main()
    {
    //freopen("treasure.in","r",stdin);
    //freopen("treasure.out","w",stdout);
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
    for(int j=1;j<=m;j++)
    {
    cin>>b[i][j]>>a[i][j];
    if(b[i][j]==1)c[i]=c[i]+1;
    }
    }
    cin>>k;l=k;
    for(int i=1;i<=n;i++)
    {
    t=0;k=l;

    ans=ans+a[i][k+1];
    if(i==n)break;
    d=a[i][k+1]%c[i];
    if(d==0)
    {
    k=k-1;if(k<0)k=m-1;
    while(b[i][k+1]==0)
    {
    k--;
    if(k<0)k=m-1;
    }
    l=k;
    }
    if(b[i][k+1]==1)t=1;
    while(t<d)
    {
    l++;
    if(l==m)l=0;
    if(b[i][l+1]==1)t++;
    }
    }
    cout<<ans%20123;
    }

  • 0
    @ 2017-11-04 20:07:49

    //从0开始!!!
    var a,b:array[0..10005,0..105]of longint;
    c:array[0..10005]of longint;
    i,j,k,n,m,p,x,ans:longint;
    begin
    readln(n,m);
    for i:=0 to n-1 do
    for j:=0 to m-1 do
    begin
    readln(a[i,j],b[i,j]);
    inc(c[i],a[i,j]);
    end;
    readln(x);
    for i:=0 to n-1 do
    begin
    ans:=(ans+b[i,x])mod 20123;
    p:=b[i,x]mod c[i];
    if p=0 then p:=c[i];
    k:=x;
    if a[i,k]=1 then dec(p);
    while p>0 do
    begin
    inc(k);
    k:=k mod m;
    if a[i,k]=1 then dec(p);
    end;
    x:=k;
    end;
    ans:=ans mod 20123;
    writeln(ans);
    end.

  • 0
    @ 2017-08-22 03:04:11

    我的第40 AC 居然是 water question

    #include<iostream>
    int a[10001][102],c,i,j,k,n,m,s,x;
    bool b[10001][101];
    int main(){
        scanf("%d%d",&n,&m);
        for (i=1;i<=n;++i)
        for (j=0;j<m;++j)
        {
            scanf("%d%d",&c,&a[i][j]);
            a[i][m]+=b[i][j]=c;
        }
        scanf("%d",&k);
        for (i=1;i<=n;++i)
        {
            x=a[i][k];
            s=(s+x)%20123;
            x=(x-1)%a[i][m]+1;
            int y=0;
            while (y+=b[i][k],y<x)
            k=(k+1)%m;
        }
        printf("%d",s);
        return 0;
    }
    
  • 0
    @ 2015-12-10 19:30:33

    此题需要注意的就是细节!!我第一次没取模,第二次取错模,第三次才对orz....
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #define rep(i, x, y) for (int i = x; i <= y; ++i)

    using namespace std;

    const int maxn = 10005, maxm = 105;

    int n, m;
    int hv[maxn][maxm], val[maxn][maxm];
    bool vis[maxm];
    int q[maxm], cnt;

    void calcu(int k, int now) {
    memset(vis, 0, sizeof (vis));
    cnt = 0;
    for (int i = now; !vis[i]; i = (i + 1) % m) {
    vis[i] = true;
    if (hv[k][i]) q[cnt++] = i;
    }
    }
    int main(int argc, const char *argv[]) {
    scanf("%d %d", &n, &m);
    rep(i, 0, n - 1) rep(j, 0, m - 1) scanf("%d %d", &hv[i][j], &val[i][j]);
    int ans = 0, now;
    scanf("%d", &now);
    rep(i, 0, n - 1) {
    ans += val[i][now];
    ans %= 20123;
    calcu(i, now);
    now = q[(val[i][now] - 1) % cnt];
    }
    printf("%d\n", ans);
    return 0;
    }

    • @ 2015-12-10 19:31:48

      顺便说一句,为什么最近Vj测评机总是挂......

  • 0
    @ 2015-10-31 11:59:29

    #include<iostream>
    #include<cstdio>
    using namespace std;
    const int MAXN = 10000 + 10;
    const int MAXM = 100 + 10;
    const int mod = 20123;

    int can[MAXN][MAXM], pos[MAXN][MAXM], quan[MAXN];

    int main()
    {
    int n, m, now;
    long long ans = 0;
    scanf("%d%d", &n, &m);
    for(int i=1; i<=n; i++)
    for(int j=0; j<m; j++){
    scanf("%d%d", &can[i][j], &pos[i][j]);
    if(can[i][j]) quan[i]++;
    }
    scanf("%d", &now);
    for(int i=1; i<=n; i++){
    ans = (ans + pos[i][now])%mod;
    int bri = 0, qiao;
    while(pos[i][now] > quan[i])
    pos[i][now] -= quan[i];
    for(int j=now; bri<pos[i][now]; j=(j+1)%m)
    if(can[i][j])
    bri++, qiao = j;
    now = qiao;
    }
    printf("%lld", ans);

    return 0;
    }

  • 0
    @ 2015-06-01 14:06:48

    program treasure;
    Var
    th:array[1..10000,1..100]of longint;
    tl:array[1..10000,1..100]of 0..1;
    xx:array[1..10000]of longint;
    n,m,i,j,sum,k:longint;
    procedure shang(x,y:longint);
    var
    a:longint;
    begin
    if(x=n+1)then exit;
    a:=th[x,y];
    sum:=(sum+a)mod 20123;
    for j:=1 to m do
    begin
    th[x,j]:=th[x,j] mod xx[x];
    if(th[x,j]=0)then th[x,j]:=xx[x];
    end;
    a:=th[x,y];
    while(a<>0)do
    begin
    if(tl[x,y]=1)then dec(a);
    if(a=0)then break;
    inc(y);
    if(y>m)then y:=y mod m;
    end;
    shang(x+1,y);
    end;
    Begin
    fillchar(xx,sizeof(xx),0);
    readln(n,m);
    sum:=0;
    for i:=1 to n do
    begin
    for j:=1 to m do
    begin
    read(tl[i,j],th[i,j]);
    if(tl[i,j]=1)then inc(xx[i]);
    end;
    end;
    read(k);
    inc(k);
    shang(1,k);
    writeln(sum);
    End.

  • 0
    @ 2014-10-25 19:10:54

    大部分时间都是花在读入上,getchar()大法好 //orzn+e
    int F(int &n)
    {
    char ch,f=0;
    while(((ch=getchar())<'0' || ch>'9') && ch!='-') ;
    ch=='-'?n=f=0:(n=ch-'0',f=1);
    while((ch=getchar())>='0'&&ch<='9') n*=10,n+=ch-'0';
    return f?n:n=-n;
    }
    直接scanf
    测试数据 #0: Accepted, time = 0 ms, mem = 9952 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 9944 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 9944 KiB, score = 10
    测试数据 #3: Accepted, time = 46 ms, mem = 9948 KiB, score = 10
    测试数据 #4: Accepted, time = 39 ms, mem = 9948 KiB, score = 10
    测试数据 #5: Accepted, time = 483 ms, mem = 9944 KiB, score = 10
    测试数据 #6: Accepted, time = 561 ms, mem = 9944 KiB, score = 10
    测试数据 #7: Accepted, time = 546 ms, mem = 9948 KiB, score = 10
    测试数据 #8: Accepted, time = 748 ms, mem = 9948 KiB, score = 10
    测试数据 #9: Accepted, time = 748 ms, mem = 9948 KiB, score = 10
    Accepted, time = 3171 ms, mem = 9952 KiB, score = 100
    优化后
    测试数据 #0: Accepted, time = 0 ms, mem = 9944 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 9948 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 9952 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 9952 KiB, score = 10
    测试数据 #4: Accepted, time = 15 ms, mem = 9948 KiB, score = 10
    测试数据 #5: Accepted, time = 78 ms, mem = 9948 KiB, score = 10
    测试数据 #6: Accepted, time = 46 ms, mem = 9952 KiB, score = 10
    测试数据 #7: Accepted, time = 93 ms, mem = 9944 KiB, score = 10
    测试数据 #8: Accepted, time = 62 ms, mem = 9944 KiB, score = 10
    测试数据 #9: Accepted, time = 93 ms, mem = 9948 KiB, score = 10
    Accepted, time = 387 ms, mem = 9952 KiB, score = 100

  • 0
    @ 2014-08-11 17:44:36

    program p1787;
    var
    a:array[0..10000,0..100,1..2]of longint;
    b:array[0..100]of longint;
    i,j,c,d,n,m,k,now,sum:longint;
    begin
    read(n,m);
    for i:=1to n do
    for j:=0to m-1do
    read(a[i,j,1],a[i,j,2]);
    read(now);
    for i:=1to n do
    begin
    sum:=(sum+a[i,now,2])mod 20123;
    d:=a[i,now,2];
    k:=0;
    for j:=now to m-1do
    if a[i,j,1]=1then
    begin
    inc(k);
    b[k]:=j;
    end;
    for j:=0to now-1do
    if a[i,j,1]=1then
    begin
    inc(k);
    b[k]:=j;
    end;
    d:=d mod k;
    if d=0then
    d:=k;
    now:=b[d];
    end;
    write(sum);
    end.

  • 0
    @ 2014-06-07 23:40:21

    C++党请注意本题数据太大使用cin会超时,请使用scanf

    劝告:NOIP、NOI中遇到大数据请使用fscanf和fprintf

  • 0
    @ 2013-10-20 16:56:39

    AC 相信党代表
    program nihaowoshituerbi;
    var
    a:array[0..10000,0..100,1..2] of longint;
    b:array[0..100] of longint;
    i,j,c,d,n,m,k,now,sum:longint;
    begin
    read(n,m);
    for i:=1 to n do
    for j:=0 to m-1 do
    read(a[i,j,1],a[i,j,2]);
    read(now);

    for i:=1 to n do
    begin
    sum:=(sum+a[i,now,2]) mod 20123;
    d:=a[i,now,2];
    k:=0;
    for j:=now to m-1 do
    if a[i,j,1]=1 then begin inc(k);b[k]:=j;
    end;
    for j:=0 to now-1 do
    if a[i,j,1]=1 then begin inc(k);b[k]:=j;
    end;
    d:=d mod k;
    if d=0 then d:=k;
    now:=b[d];

    end;
    write(sum);
    end.

  • 0
    @ 2013-08-21 14:40:42

    比赛的时候写过了
    主要注意取余 否则会超时

  • 0
    @ 2013-08-13 14:45:44

    刚才的不对,修改了一下:
    #include <iostream>
    using namespace std;
    int a[10005][105],b[10005][105],shu[10005];
    int main()
    {
    int i,j,k,n,m,c,x,ans=0;
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++){
    for(j=0;j<m;j++){
    scanf("%d%d",&a[i][j],&b[i][j]);
    shu[i]+=a[i][j];
    }
    }
    scanf("%d",&x);
    for(i=0;i<n;i++){
    ans+=b[i][x];
    ans%=20123;
    c=b[i][x]%shu[i];
    if(c==0)c=shu[i];k=x;
    if(a[i][k]==1)c--;
    while(c>0){
    k++;
    k=k%m;
    if(a[i][k]==1)c--;

    }
    x=k;
    }
    ans%=20123;
    printf("%d",ans);
    printf("\n");
    return 0;
    }

  • 0
    @ 2013-08-13 14:44:39

    #include <iostream>
    using namespace std;
    int a[10005][105],b[10005][105],shu[10005];
    int main()
    {
    int i,j,k,n,m,c,x,ans=0;
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++){
    for(j=0;j<m;j++){
    scanf("%d%d",&a[i][j],&b[i][j]);
    shu[i]+=a[i][j];
    }
    }
    scanf("%d",&x);
    for(i=0;i<n;i++){
    ans+=b[i][x];
    ans%=20123;
    c=b[i][x]%shu[i];
    if(c==0)c=shu[i];k=x;
    if(a[i][k]==1)c--;
    while(c>0){
    k++;
    k=k%m;
    if(a[i][k]==1)c--;

    }
    x=k;
    }
    ans%=20123;
    printf("%d",ans);
    printf("\n");

    system("pause");
    return 0;
    }
    (C++)

  • 0
    @ 2013-07-04 16:57:35

    看下那个地方错了,大婶们
    type ds=record
    f:0..1;
    x:longint;
    end;
    var n,m,b:integer;
    a:array[1..10000,0..99]of ds;
    c:array[1..10000]of integer;
    t:longint;
    procedure init;
    var i,j,s:integer;
    begin
    readln(n,m);
    for i:=1 to n do
    begin
    s:=0;
    for j:=0 to m-1 do
    begin
    readln(a[i,j].f,a[i,j].x);
    s:=s+a[i,j].f;
    end;
    c[i]:=s;
    end;
    readln(b);
    t:=0;
    end;
    procedure out;
    begin
    writeln(t);
    end;
    procedure work(floor,room:integer);
    var i,j,k,tc:longint;
    begin
    i:=floor;j:=room;
    k:=a[i,j].x;t:=(t+k) mod 20123;
    k:=k mod c[i];
    tc:=a[i,j].f;
    while tc<k do
    begin
    inc(j);
    if j=m then j:=0;
    tc:=tc+a[i,j].f;
    end;
    if i<n then work(i+1,j);
    end;
    begin
    init;
    work(1,b);
    out;
    end.

  • 0
    @ 2012-11-24 17:41:23

    答案错误求改

    program P1787;

    var

    n,m,i,j,l,k,now:integer;

    s:longint;

    a,b,c,d:array[1..10000,0..100] of longint;

    e:array[1..10000] of integer;

    begin

    read(n,m);

    for i:=1 to n do   begin

    for j:=0 to m-1 do begin

    read(a,b);

    if a=1 then begin e[i]:=e[i]+1; k:=k+1; c:=j; end;

    end;

    k:=0;

    end;

    readln(l);

    now:=l;

    s:=s+b[1,now];

    for i:=1 to n-1 do begin

    for j:=l to l+m do

    if a[i,j mod m]=1 then begin l:=j mod m; break;end;

    for j:=1 to e[i] do

    if c=l then begin l:=j; break; end;

    l:=(l+b-a) mod e[i];

    now:=c;

    s:=s+b;

    l:=now;

    end;

    write(s mod 20123);

    end.

    全错

    • @ 2013-10-20 17:14:15

      if k=0 then ......

  • 0
    @ 2012-11-22 17:13:42

    program treasure;

    var

    s:array[0..10000,0..100,1..2] of longint;

    x:array[0..100] of longint;

    a,b,c,d,n,m,k,now,sum:longint;

    begin

    {assign(input,'treasure.in');

    reset(input);

    assign(output,'treasure.out');

    rewrite(output);}

    read(n,m);

    for a:=1 to n do

    for b:=0 to m-1 do

    read(s[a,b,1],s[a,b,2]);

    read(now);

    for a:=1 to n do

    begin

    sum:=(sum+s[a,now,2]) mod 20123;

    d:=s[a,now,2];

    k:=0;

    for b:=now to m-1 do

    if s[a,b,1]=1 then begin inc(k);x[k]:=b;end;

    for b:=0 to now-1 do

    if s[a,b,1]=1 then begin inc(k);x[k]:=b;end;

    d:=d mod k;

    if d=0 then d:=k;

    now:=x[d];

    end;

    write(sum);

    {close(input);

    close(output);}

    end.

  • 0
    @ 2012-11-21 20:06:39

    ├ 测试数据 01:答案正确... (0ms, 8600KB)

    ├ 测试数据 02:答案正确... (15ms, 8600KB)

    ├ 测试数据 03:答案正确... (15ms, 8600KB)

    ├ 测试数据 04:答案正确... (15ms, 8600KB)

    ├ 测试数据 05:答案正确... (31ms, 8600KB)

    ├ 测试数据 06:答案正确... (323ms, 8600KB)

    ├ 测试数据 07:答案正确... (382ms, 8600KB)

    ├ 测试数据 08:答案正确... (383ms, 8600KB)

    ├ 测试数据 09:答案正确... (416ms, 8600KB)

    ├ 测试数据 10:答案正确... (426ms, 8600KB)

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

    Accepted / 100 / 2011ms / 8600KB

    ├ 测试数据 01:答案正确... (102ms, 672KB)

    ├ 测试数据 02:答案正确... (0ms, 672KB)

    ├ 测试数据 03:答案正确... (13ms, 672KB)

    ├ 测试数据 04:答案正确... (0ms, 672KB)

    ├ 测试数据 05:答案正确... (15ms, 672KB)

    ├ 测试数据 06:答案正确... (15ms, 672KB)

    ├ 测试数据 07:答案正确... (13ms, 672KB)

    ├ 测试数据 08:答案正确... (15ms, 672KB)

    ├ 测试数据 09:答案正确... (15ms, 672KB)

    ├ 测试数据 10:答案正确... (15ms, 672KB)

    注意mod出0

    还是没能秒杀啊

  • -1
    @ 2017-07-15 20:39:45

    额,对是对了,就是不知道为啥当mod得出0时为啥会这样处理的
    Var
    n,m,i,j,sum,num,l1,k,wz,k1:longint;
    a,b:array[1..10000,0..100]of longint;
    f:array[1..10000,1..100]of longint;
    l:array[1..10000]of longint;
    Begin
    readln(n,m);
    for i:=1 to n do
    for j:=0 to m-1 do
    begin
    read(a[i,j]);
    if a[i,j]=1 then
    begin
    inc(l[i]);
    f[i,l[i]]:=j;
    end;
    readln(b[i,j]);
    end;
    readln(k);
    l1:=1;
    while l1<=n do
    begin
    num:=(num+b[l1,k]) mod 20123;
    if b[l1,k] mod l[l1]=0 then
    sum:=l[l1]
    else
    sum:=b[l1,k] mod l[l1];
    wz:=0;
    while wz<sum do
    begin
    if a[l1,k]=1 then
    inc(wz);
    if k=m-1 then
    k:=0
    else
    inc(k);
    end;
    if k=0 then
    k:=m-1
    else
    dec(k);
    inc(l1);
    end;
    writeln(num);
    readln;
    End.

信息

ID
1787
难度
6
分类
模拟 点击显示
标签
递交数
2633
已通过
612
通过率
23%
被复制
10
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