RQNOJ 前100题 -练手
1.http://www.rqnoj.cn/problem/1
program mingming;
var
a:array [1..1000] of integer;
i,j,k,n:longint;
begin
fillchar(a,0,sizeof(a));
k:=0;
read(n);
for i:=1 to n do
begin
read(j);
if a[j] <> 1 then
begin
a[j]:=1;
inc(k);
end;
end;
writeln(k);
for i:=1 to 1000 do
begin
if a[i] = 1 then
write(i,' ');
end;
end.
2.
#include <stdio.h>
#define maxm 26
#define maxn 30001
int n,m;
int v[maxm],w[maxm];
int f[maxn]={0};
int main(int argc, const char *argv[])
{
// freopen("1.in","r",stdin);
int i,j,k,l;
scanf("%d%d",&n,&m);
for(i=1;i<=m;i++)
scanf("%d%d",&v[i],&w[i]);
for(i=1;i<=m;i++)
for(j=n;j>0;j--)
{
if(j>=v[i] && (f[j] < f[j-v[i]] + v[i]*w[i]))
f[j] = f[j- v[i]] + v[i]*w[i];
}
printf("%d",f[n]);
return 0;
}
3.
#include <stdio.h>
#include <stdlib.h>
char JamNum[27]={0};
char Snum[27]= {0};
int weishu,Start,End;
int main()
{
char Schar,Echar,temp;
int change;
int i,j,k,l;
scanf("%d%d%d",&Start,&End,&weishu);
scanf("%s",Snum);
//printf("%s",Snum);
Schar=Start+'a'-1;Echar=End+'a'-1;
//printf("%c %c",Schar,Echar);
for(j=0;j<5;j++)
{
change = 26;
for(i=weishu-1;i>=0;i--)//判断那一位可以改
{
if(Snum[i]<Echar+i-weishu+1)
{
change=i;
break;
}
}
if(change == 26)
return 0;
temp=Snum[change];
for(i=change;i<weishu;i++)
{
Snum[i]=++temp;
}
printf("%s\n",Snum);
}
return 0;
}
4.
#include <stdio.h>
#define maxn 1000
int f[1001]= {1,1};
int k,N;
int pow(int a,int b)
{
int r= 1,base = a;
while(b != 0)
{
if(b&1)
r*= base;
base *= base;
b>>=1;
}
return r;
}
int main(int argc, const char *argv[])
{
int i,j,l,temp,temp2;
scanf("%d%d",&k,&N);
l=1;
j=2;
// temp = pow(N,l);
temp =2;
for(i=2;i<N+1;i++)
{
if(j < temp ){
f[i] =temp2 + f[j++];
}
else
{
temp2 = f[i] = pow(k,l++);
j=1,temp = i;
}
}
printf("%d",f[N]);
return 0;
}
5.
/*
* =====================================================================================
*
* Filename: pid5.c
*
* Description: PID5/能量项链
*
* Version: 1.0
* Created: 2014-9-23 17:31:43
* Revision: none
* Compiler: gcc
*
* Author: Rainboy (mn), 597872644@qq.com
* Company: NONE
*
* =====================================================================================
*/
#include <stdio.h>
#define maxn 1000
int f[301][301]= {0};
int N;
int u[301],v[301];
int main(int argc, const char *argv[])
{
freopen("energy.in","r",stdin);
freopen("energy.out","w",stdout);
int i,j,k,l;
scanf("%d",&N);
scanf("%d",&i);
u[1] = i; v[N] = i;
for(i=2;i<N+1;i++)
{
scanf("%d",&j);
u[i]= j; v[i-1]=j;
}
for(i=N+1;i<2*N;i++)
{
u[i]= u[i-N];
v[i] = v[i-N];
}
for(i=1;i<N+1;i++)
for(l=1;l<N;l++)//宽度
for(j=i;j<N+i;j++)
for(k=j;k+1<=j+l;k++)
{
if(f[j][j+l] < f[j][k]+f[k+1][j+l] + u[j]*v[k]*v[j+l])
f[j][j+l] = f[j][k]+f[k+1][j+l] + u[j]*v[k]*v[j+l];
}
l = 0;
for(i=1;i<N+1;i++)
if(l<f[i][i+N-1] ) l=f[i][i+N-1];
printf("%d",l);
return 0;
}
6.
/*
* =====================================================================================
*
* Filename: 6.c
*
* Description: 金明的预算方案
*
* Version: 1.0
* Created: 2014-9-24 12:37:55
* Revision: none
* Compiler: gcc
*
* Author: Rainboy (mn), 597872644@qq.com
* Company: NONE
*
* =====================================================================================
*/
#include <stdio.h>
int N,m;
#define maxn 200
//有附件的只是有多种选择而已，不买，买主件，买主件+附件1，买主件+附件2，等等 共有4种可能性
//所以我们只处理我们的数据结构就行了
//每个主件最多有2个附件 让这一题的难度降低很多
int f[maxn][maxn] ={0};
int val[maxn][maxn] = {0};
int ans[32001]= {0};
int main(int argc, const char *argv[])
{
int i,j,k,l,v,p,q;
scanf("%d%d",&N,&m);
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&v,&p,&q);
if(q == 0)
{
f[i][1] = v;
val[i][1] = p;
++f[i][0];
}
else
{
f[q][++f[q][0]]= v;
val[q][f[q][0]] = p;
}
}
for(i=1;i<=m;i++)
{
if(f[i][0] !=0 )
{
for(j=N;j>0;j--) // 判断 附件有多少 0-1-2
{
if(j-f[i][1] >=0 && ans[j] < f[i][1]*val[i][1] + ans[(j-f[i][1])] ) ans[j] = f[i][1]*val[i][1] + ans[(j-f[i][1])];// 只买主件
if(f[i][0] >1)// 买主件 + 附件1
{
if(j-f[i][1]-f[i][2] >=0 && ans[j] < f[i][1]*val[i][1] + f[i][2]*val[i][2] +ans[(j-f[i][1]-f[i][2])])
ans[j] = f[i][1]*val[i][1] + f[i][2]*val[i][2] +ans[(j-f[i][1]-f[i][2])];
if(f[i][0] == 3)
{//买附件2 + 主件
if(j-f[i][1] - f[i][3] >=0 && ans[j] < f[i][1]*val[i][1] + f[i][3]*val[i][3] + ans[(j-f[i][1] - f[i][3])])
ans[j] < f[i][1]*val[i][1] + f[i][3]*val[i][3] + ans[(j-f[i][1] - f[i][3])] ;
//买附件1+2+主件
if(j-f[i][1] - f[i][2] -f[i][3] >= 0 && ans[j] < f[i][1]*val[i][1] + f[i][2] * val[i][2] + f[i][3]*val[i][3] + \
ans[(j-f[i][1] - f[i][2] - f[i][3])])
ans[j] = f[i][1]*val[i][1] + f[i][2] * val[i][2] + f[i][3]*val[i][3] + \
ans[(j-f[i][1] - f[i][2] - f[i][3])];
}
}
}
}
}
printf("%d",ans[N]);
return 0;
}
7. 题目的描述真的太长了，这个题目的难度就在于他的长度。看来功夫在诗外啊，如果NOIP出一题阅读时间超长的题，30分钟，绝对是前10难度；
/*
* =====================================================================================
*
* Filename: 7-2.c
*
* Description: PID7
*
* Version: 1.0
* Created: 2014-9-25 16:48:22
* Revision: none
* Compiler: gcc
*
* Author: Rainboy (mn), 597872644@qq.com
* Company: NONE
*
* =====================================================================================
*/
#include <stdio.h>
int jiqi[21][21];// 第i 个工件 第j步用的机器
int gongxu[500]; // 判断工件 是哪个工序
int gongxuf[500]= {0};// 辅助
int time[21][21];// 工序用的时间
int xl[500];// 输入的工件序列
int bool[21][1000] = {0};
int finishTime[21] ={0}; // 工件 前一个工序结束的时间
int n,m;
int main(int argc, const char *argv[])
{
int i,j,k,l;
int t,temp,mc;
scanf("%d%d",&m,&n);//m-机器数 n-工件数
for(i=1;i<=m*n;i++)
{
scanf("%d",&j);
xl[i] = j;
gongxu[i] = ++gongxuf[j];
}
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
scanf("%d",&jiqi[i][j]);
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
scanf("%d",&time[i][j]);
for(i=1;i<=m*n;i++)
{
t = xl[i];// t 表示现在要加入的工件的编号
j = finishTime[t]; // j 表示工件上一步结束的时间
mc = jiqi[t][gongxu[i]];// mc 表示 用到的机器
// 下面的就是进行 见缝插针
temp = 0;
for(k=j+1;;k++)
{
if(bool[mc][k] == 0)
{
if( ++ temp == time[t][gongxu[i]])
{
bool[mc][0] = bool[mc][0] > k ? bool[mc][0]: k;
finishTime[t] = k;
for(l=k- time[t][gongxu[i]] +1;l<=k;l++)
bool[mc][l] = 1;
break;
}
}
else
temp = 0;
}
}
temp =0;
for(i=1;i<=m;i++)
if(temp < bool[i][0])
temp = bool[i][0];
printf("%d",temp);
return 0;
}
