九度oj 专题整理 （模拟部分1）

分类参考链接    点击打开链接

假期陆续刷了很多水题，觉得ACM只要能理解思路的代码都是水题，套路也是有的，不过我的水平也就停留在水题上了，随着面试日期的临近，反而今天有些不静心，就不如现在整理一下自己做过的水题吧，也免得眼高手低，现在看看自己当时写的代码，很多思路也让现在的自己吃惊。因为自己总是喜欢倒序做事，那就从每个专题的题目倒序来吧

模拟类型：1000 1001 1020 1031 1036 1038 1013 1014 1045 1046 1048 1050 1059 1060 1062 1063 1064 1065 1067 1068 1070 1075  1177 1179 1183 1186

题目：1186（简单来说，就是判断闰年）因为之前做过类似的题目，这个就不再写一遍了

题目 1 Maya Calendar 点击打开链接

题目2 打印月历 点击打开链接

1.Maya Calendar，这个题目如果先加上1，再去对月份和星期取余，换过各种方式，样例过了，却总WA,现在还不懂为什么，所以参照网上的，改成先取余后加了，

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<queue>#include<set>#include<map>#define LL long long#define PI acos(-1)#define exp 1e-9using namespace std;priority_queue <int, vector<int>, greater<int> > Q;char Haab[20][20]={"pop", "no", "zip", "zotz", "tzec", "xul", "yoxkin", "mol", "chen", "yax", "zac", "ceh", "mac", "kankin", "muan","pax", "koyab", "cumhu", "uayet"};char holly[20][20]={"imix", "ik", "akbal", "kan", "chicchan", "cimi", "manik", "lamat", "muluk", "ok", "chuen", "eb", "ben", "ix", "mem", "cib", "caban", "eznab", "canac", "ahau"};int main(){    //printf("%s",Haab[1]);    int n;    scanf("%d",&n);    //printf("%d\n",n);    int day,year;    char month[10];    int sum[10000];    memset(sum,0,sizeof(sum));    for(int j=0; j<n; j++)    {        scanf("%d.%s %d",&day,month,&year);        sum[j]+=year*365;        for(int i=0; i<19; i++)        {            if(strcmp(Haab[i],month)==0)            {                sum[j]+=i*20;                break;            }        }        sum[j]+=day;        //printf("%d.%s %d",day,month,year);    }    printf("%d\n",n);    //printf("%.8lf\n",16/10);    for(int k=0; k<n; k++)    {        int y=sum[k]/260;        //printf("%d\n",sum[k]);        int m=sum[k]%20;        int d=sum[k]%13+1;        printf("%d %s %d\n",d,holly[m],y);    }    return 0; }

2.打印月历

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#define exp 1e-6#define pi acos(-1.0)using namespace std;int m1[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};int m2[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};bool judge(int year){    if(year%100==0)    {        if(year%400==0)        {            return true;        }    }    else    {        if(year%4==0)        {            return true;        }    }    return false;}int daynum(int year){    int day=0;    for(int i=1900; i<year; i++)    {        if(judge(i))        {            day+=366;        }        else        {            day+=365;        }    }    return day;}int monthday(int year,int month){    int mday=0;    if(judge(year))    {        for(int i=0; i<month; i++)        {            mday+=m2[i];        }    }    else    {        for(int i=0; i<month; i++)        {            mday+=m1[i];        }    }    return mday;}int main(){    int year,month;    while(~scanf("%d %d",&year,&month))    {        int day=daynum(year);        day+=monthday(year,month);        printf("Sun Mon Tue Wed Thu Fri Sat\n");        int flag=0;        int mod=day%7+1;        for(int k=0; k<mod; k++)        {            printf("    ");            flag++;        }        if(judge(year))        {            for(int i=1; i<m2[month];i++)            {                printf("%3d ",i);                flag++;                if(flag%7==0)                printf("\n");            }            printf("%3d\n",m2[month]);        }        else        {            for(int i=1; i<m1[month];i++)            {                printf("%3d ",i);                flag++;                if(flag%7==0)                printf("\n");            }            printf("%3d\n",m1[month]);        }       // printf("%d\n",day%7);    }    return 0;}

题目1183：守形数

现在看上去的想法还可以，大概是钻了数很小的空子，不然别的思路呢？

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<queue>#include<set>#include<map>#define LL long long#define PI acos(-1)#define exp 1e-9using namespace std;priority_queue <int, vector<int>, greater<int> > Q;  int main(){    int n,m,ans;    while(~scanf("%d",&n))    {        m=n*n;        if(n<10)            ans=m%10;        else if(n<100)            ans=m%100;        else            ans=m%1000;        if(ans==n)            printf("Yes!\n");        else            printf("No!\n");    }    return 0; } 

题目1179：阶乘，

题目很简单，可能对于初学者来说，就是要注意阶乘数据类型要用long long 吧，还做过一个题目要用double，发现自己之前wa过两次，原来将n%2判断奇数偶数写成了n%1 = =

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<queue>#include<set>#include<map>#define LL long long#define PI acos(-1)#define exp 1e-9#define INF 0x7fffffffusing namespace std;priority_queue <int, vector<int>, greater<int> > Q;long long list[1010];int n;int fun(){    list[0]=1;    for(int i=1; i<=n; i++)    {        list[i]=list[i-1]*i;        //printf("%d\n",list[i]);    } }int main(){    while(~scanf("%d",&n))    {        fun();        LL ans=0,cnt=0;        if(n%2)        {            for(int i=1; i<=n; i+=2)            {                ans+=list[i];            }            for(int i=2; i<=n-1; i+=2)            {                cnt+=list[i];            }        }        else        {            for(int i=1; i<=n-1; i+=2)            {                ans+=list[i];            }            for(int i=2; i<=n; i+=2)            {                cnt+=list[i];            }        }        printf("%lld %lld\n",ans,cnt);    }    return 0;}

太长了，三个题为单位续写吧，，