今天晚上本希望参加提高组的模拟,但是提高组的发题的人迟到进3个小时,时间就是金钱啊,再说那样就太晚了,只得去参加今晚的普及组模拟赛,题目简单的不能再简单了,1个小时就搞定了

我的代码:

buy.c

#include<stdio.h>
int main(void)
{
    int i,j,k;
    int x,v,temp;
    int c,n;
    int countline[10001]={0};
    FILE *input,*output;
    input=fopen("buy.in","r");
    utput=fopen("buy.out","w");
    fscanf(input,"%d%d",&c,&n);
    for(i=0;i<n;i++)
    {
      fscanf(input,"%d%d",&x,&v);
      for(j=c;j>=x;j--)
      {
        if(countline[j]<(temp=(countline[j-x]+v)))
          countline[j]=temp;
      }
    }
    fprintf(output,"%d\n",countline[c]);
    fclose(input);
    fclose(output);
    return 0;
}
day.c

#include<stdio.h>
int main(void)
{
    int i,j,k,x,y,z;
    int m[12]={31,28,31,30,31,30,31,31,30,31,30,31};
    int y1,y2,m1,m2,d1,d2;
    int sum=0;
    FILE *input,*output;
    input=fopen("day.in","r");
    utput=fopen("day.out","w");
    fscanf(input,"%d%d%d%d%d%d",&y2,&m2,&d2,&y1,&m1,&d1);
    for(i=y1;i<y2;i++)
    {
      if(i%4) x=365; else x=366;
      sum+=x;
    }
    for(i=0;i<m2-1;i++)
      sum+=m[i];
    if(y2%4==0 && m2>2) sum++;
    sum+=d2;
    for(i=0;i<m1-1;i++)
      sum-=m[i];
    if(y1%4==0 && m1>2) sum--;
    sum-=d1;
    fprintf(output,"%d\n",sum);
    fclose(input);
    fclose(output);
    return 0;
}
   
present.c

#include<stdio.h>
typedef struct
{
  int num; /*编号*/
  int x; /*分数*/
} DATA;
DATA data[100];
int x,y;
int n;
FILE *input,*output;
/*Qsort!*/
void qsort(int low,int high,DATA key[])
{
     int i=low,j=high;
     DATA tag;
     if(i<j)
     {
       tag=key[i];
       do
       {
         while(((tag.x>key[j].x)||(tag.x==key[j].x && tag.num<=key[j].num)) && i<j) j--;
         if(i<j)
         {
           key[i]=key[j];
           i++;
           while(((tag.x<key[i].x)||(tag.x==key[i].x && tag.num>key[i].num)) && i<j) i++;
           if(i<j)
           {
             key[j]=key[i];
             j--;
           }
         }
       }while(i<j);
       key[i]=tag;
       qsort(low,j-1,key);
       qsort(i+1,high,key);
     }
}
/*I want to fly higher!*/
int main(void)
{
    int i,j,k;
    input =fopen("present.in","r");
    utput=fopen("present.out","w");
    fscanf(input,"%d",&n);
    for(i=0;i<n;i++)
      fscanf(input,"%d%d",&data[i].num,&data[i].x);
    qsort(0,n-1,data);
    for(i=0;i<n;i++)
      fprintf(output,"%d %d %d\n",i+1,data[i].num,data[i].x);
    fclose(input);
    fclose(output);
    return 0;
}
save.c

#include<stdio.h>
int main(void)
{
    int i,j,k;
    int n;
    int x,y;
    int sum=0;
    FILE *input,*output;
    input=fopen("save.in","r");
    utput=fopen("save.out","w");
    fscanf(input,"%d",&n);
    for(i=0;i<n;i++)
    {
      fscanf(input,"%d%d",&x,&y);
      sum=sum+x-y;
    }
    if(sum>0)
    {
      fprintf(output,"Yes\n%d\n",sum);
    }
    else
    {
      fprintf(output,"No\n");
    }
    fclose(input);
    fclose(output);
    return 0;
}

2关键字排序+DP基础(0/1背包)+简单模拟

数据规模相当小,搜有时可能都比DP快...