POJ

1003  Hangover

#include<stdio.h>
int main()
{
    double c, cur_len;
    scanf_s("%lf", &c);
    while (1)
    {
        if (c < 0.0001)
            break;
        cur_len = 1.0 / 2;
        int count = 1;
        while (cur_len<c)
        {
            count++;
            cur_len += 1.0 / (count + 1);
        }
        printf("%d card(s)\n", count);
        scanf_s("%lf", &c);
    }
    return 0;
}

1004  Financial Management

/*
    水题，求平均数
    注意小数点后保留两位即可
*/
#include<stdio.h>
int main()
{    
    int i;
    double sum=0, num,result;
    for (i = 0; i < 12; i++)
    {
        scanf_s("%lf", &num);
        sum += num;
    }
    printf("$%.2lf\n", sum / 12);
    return 0;
}

1005  I Think I Need a Houseboat

/*
    注意每年累加的是面积，不是半径，半径会越来越小
*/
#include<stdio.h>
#include<math.h>
#define pi 3.141592
int main()
{
    int N;
    double X, Y,Z;
    scanf_s("%d", &N);
    int count = 1;
    while (N--)
    {
        scanf_s("%lf%lf", &X, &Y);
        Z = pi*(X*X + Y*Y) / 100;
        printf("Property %d: This property will begin eroding in year %d.\n", count, int(Z+1));
        ++count;
    }
    printf("END OF OUTPUT.\n");
    return 0;
}

1006  Biorhythms

#include<iostream>
using namespace std;
int main()
{    int p,e,i,d,j,no=1;
    cin>>p>>e>>i>>d;
    while(p!=-1 && e!=-1 && i!=-1 && d!=-1)
    {    for(j=d+1;j<=21252;j++)
            if((j-p)%23==0)
                break;
        for(;j<=21252;j=j+23)
            if((j-e)%28==0)
                break;
        for(;j<=21252;j=j+23*28)
            if((j-i)%33==0)
                break;
        cout<<"Case "<<no;
        cout<<": the next triple peak occurs in ";
        cout<<j-d<<" days."<<endl;
        cin>>p>>e>>i>>d;
        no++;
    }
    return 0;
}

1007  DNA Sorting

#include<iostream>
#include<string>
using namespace std;
struct dna
{
    int num;
    string s;
}DNA[100];
int inversionNum(string s)
{
    int ans = 0,len;
    int A, C, G;
    A = C = G = 0;
    len = s.length();
    for (int i = len - 1; i >= 0; i--)
    {
        switch (s[i])
        {
        case 'A':
            A++;
            break;
        case 'C':
            C++;
            ans += A;
            break;
        case 'G':
            G++;
            ans += A;
            ans += C;
            break;
        case 'T':
            ans += A;
            ans += C;
            ans += G;
        }
    }
    return ans;
}
int cmp(const void *a, const void *b)
{
    return (*(dna*)a).num > (*(dna*)b).num ? 1 : -1;
}
int main()
{
    int n, m, i;
    cin >> n >> m;
    for (i = 0; i < m; i++)
    {
        cin >> DNA[i].s;
        DNA[i].num = inversionNum(DNA[i].s);
    }
    qsort(DNA,m,sizeof(DNA[0]),cmp);
    for (i = 0; i < m; i++)
        cout << DNA[i].s << '\n';
    return 0;
}

1008  Maya Calendar

#include<iostream>
#include<string>
using namespace std;
string m[19] = {"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax",
                "zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"};
string d[20] = {"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk","ok",
                "chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};
int search(string hm)
{
    for (int i = 0; i < 19; i++)
        if (hm == m[i])
            return i;
}
int main()
{
    int n,hd,hy,hm1,hdays,ty,days;
    string hm;
    cin >> n;
    cout << n << endl;
    while (n--)
    {
        cin >> hd >> hm >> hm >> hy;
        hm1 = search(hm);
        hdays = hy * 365 + hm1 * 20 + hd ;
        ty = hdays / 260;
        days = hdays % 260;
        cout << days % 13+1 << " " << d[days % 20] << " " << ty << endl;
    }
    return 0;
}

1017  Packets

#include<iostream>
using namespace std;
int main()
{    int b1,b2,b3,b4,b5,b6; //不同大小的木块个数
    int nTotal=0; //最少需要的箱子数目
    int c1; //当前能放1*1木块的空格数目
    int c2; //当前能放2*2木块的空格数目
    int Contain2[4]={0,5,3,1};
    while(1)
    {    cin>>b1>>b2>>b3>>b4>>b5>>b6;
        if(b1==0 && b2==0 && b3==0 && b4==0 && b5==0 && b6==0)
            break;
        nTotal=b6+b5+b4+(b3+3)/4;
        c2=5*b4+Contain2[b3%4];
        if(b2>c2)    nTotal+=(b2-c2+8)/9;
        c1=36*nTotal-36*b6-25*b5-16*b4-9*b3-4*b2;
        if(b1>c1)    nTotal+=(b1-c1+35)/36;
        cout<<nTotal<<endl;
    }
    return 0;
}

1083  Moving Tables

/* 
    方法：记录每个房间前的走廊被用过多少次，答案为10*次数
    注意范围的确定，若下限为偶数，则下限需减1，若上限为奇数，则上限需加1
    例如，s=4，t=7时，房间4到房间7之间的走廊都要占用，而房间4前的走廊与房间3前的走廊是一样的，
    房间7前与房间8前的走廊也是一样的，所以要扩大的更新范围为[3，8]
*/

#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;
int cmp(const void *a, const void *b)
{
    return *(int *)a > *(int *)b ? 1 : -1;
}
int main()
{
    int T,N,s,t,i,j,temp;
    int room[400];
    scanf_s("%d", &T);
    while (T--)
    {
        memset(room, 0, sizeof(room));//注意每次测试都要初始化，而不是在输入测试次数前初始化
        scanf_s("%d", &N);
        for (i = 0; i < N; i++)
        {
            scanf_s("%d%d", &s, &t);
            if (s > t)
            {
                temp = s;
                s = t;
                t = temp;
            }
            if (s % 2 == 0)  
                s--;
            if (t % 2 != 0)
                t++;
            for (j = s; j <t; j+=2)
                room[j]++;
        }
        qsort(room, 400, sizeof(room[0]), cmp);
        printf("%d\n", 10*room[399]);
    }
    return 0;
}

1088  滑雪

#include<iostream>
using namespace std;
int map[101][101] = { 0 };//存原始数据
int r, c;//行数，列数
int dp[101][101] = { 0 };//存储节点的最长路径值
int dfs(int from, int to)
{
    //如果已经记录了值，直接返回
    if (dp[from][to] > 0)
        return dp[from][to];
    int zuo=0, you=0, shang=0, xia=0;
    //处理下路，并剪枝
    if (from + 1 <= r)
    {
        if (map[from + 1][to] < map[from][to])
            xia = 1 + dfs(from + 1, to);
        else
            xia = 1;
    }
    else
        xia = 1;
    //处理上路，并剪枝
    if (from - 1 > 0)
    {
        if (map[from - 1][to] < map[from][to])
            shang = 1 + dfs(from - 1, to);
        else
            shang = 1;
    }
    else
        shang = 1;
    //处理右路，并剪枝
    if (to + 1 <= c)
    {
        if (map[from][to + 1] < map[from][to])
            you = 1 + dfs(from, to + 1);
        else
            you = 1;
    }
    else
        you = 1;
    //处理左路，并剪枝
    if (to - 1 > 0)
    {
        if (map[from][to - 1] < map[from][to])
            zuo = 1 + dfs(from, to - 1);
        else
            zuo = 1;
    }
    else
        zuo = 1;
    //返回最大值
    if (zuo < you)
        zuo = you;
    if (zuo < shang)
        zuo = shang;
    if (zuo < xia)
        zuo = xia;
    return zuo;
}
int main()
{
    int i, j;
    cin >> r >> c;
    for (i = 1; i <= r; i++)
        for (j = 1; j <= c; j++)
            cin >> map[i][j];
    int ans = 0;
    for (i = 1; i <= r; i++)
        for (j = 1; j <= c; j++)
        {
            //记录每个节点的最长路径
            dp[i][j] = dfs(i, j);
            //选出最大长度
            if (ans < dp[i][j])
                ans = dp[i][j];
        }
    cout << ans << endl;
    return 0;
}

1258  Agri-Net （最小生成树）

#include<stdio.h>
#include<string.h>
#define MaxInt 100000
#define N 101
//创建map二维数组存储图表，low数组记录每2个点间的最小权值，
//visited数组记录某点是否已访问
int map[N][N], low[N], visited[N];
int n;
int prim()
{
    int i, j, pos, min, result=0;
    memset(visited, 0, sizeof(visited));
    //从某点开始，分别标记和记录该点
    visited[1] = 1;
    pos = 1;
    //第一次给low数组赋值
    for (i = 1; i <= n; i++)
            low[i] = map[pos][i];
    //再运行n-1次
    for (i = 1; i < n; i++)
    {
        min = MaxInt;
        //找出最小权值并记录位置
        for(j = 1; j <= n; j++)
            if (visited[j] == 0 && min > low[j])
            {
                min = low[j];
                pos = j;
            }
        //最小权值累加
        result += min;
        //标记该点
        visited[pos] = 1;
        //更新权值
        for (j = 1; j <= n; j++)
            if (visited[j] == 0 && low[j] > map[pos][j])
                low[j] = map[pos][j];
    }
    return result;
}
int main()
{
    int i, j;
    while (scanf_s("%d", &n) != EOF)
    {
        for (i = 1; i <= n; i++)
            for (j = 1; j <= n; j++)
                scanf_s("%d", &map[i][j]);
        printf("%d\n", prim());
    }
    return 0;
}

1328  Radar Installation（贪心）

#include<iostream>
#include<cmath>
#include<cstdlib>
using namespace std;
struct point
{    
    double left, right;
}p[1001];
int cmp(const void *a, const void *b)
{    
    return (*(point *)a).left> (*(point *)b).left ? 1 : -1;
}
int main()
{
    int i,n,d,count = 0,y;
    double dist,x;
    bool flag;
    while (cin >> n >> d && n != 0 && d != 0)
    {
        count++;
        flag = false;
        if (d <0)  flag = true;
        for (i = 0; i < n; i++)
        {
            cin >> x >> y;
            if (abs(y) <= d)
            {
                //计算区间
                dist = sqrt(double(d*d - y*y));
                p[i].left = x - dist;
                p[i].right = x + dist;
            }
            else
                flag = true;
        }
        if (flag)
        {
            cout << "Case " << count << ": -1"<<endl ;
            continue;
        }
        qsort(p,n,sizeof(p[0]),cmp);//排序
        x = p[0].right; 
        int m = 1;//雷达个数初始化为1
        //寻找最少区间个数
        for (i = 1; i < n; i++)
        {
            if (p[i].left > x)
            {
                m++;
                x= p[i].right;
            }
            else
            {
                if (p[i].right < x)
                    x= p[i].right;//更新右端点
            }
        }
        cout << "Case " << count << ": " << m << endl; 
    }
    return 0;
}

1331  Multiply

#include<stdio.h>
#include<string.h>
int b2ten(int x,int b);
int main()
{    int p,q,r,n,b;
    scanf("%d",&n);
    while(n--)
    {    scanf("%d%d%d",&p,&q,&r);
        for(b=2;b<=16;b++)
        {    long p2=b2ten(p,b);
            long q2=b2ten(q,b);
            long r2=b2ten(r,b);
            if(p2==-1||q2==-1||r2==-1)    continue;
            if(p2*q2==r2)    { printf("%d\n",b); break; }
        }
        if(b==17)    printf("0\n");
    }
    return 0;
}
int b2ten(int x,int b)
{    char tmp[100];
    int ret=0;
    sprintf(tmp,"%d",x);
    int len=strlen(tmp);
    for(int i=0;i<len;i++)
    {    if(tmp[i]-'0'>=b)    return -1;
        ret*=b;
        ret+=tmp[i]-'0';
    }
    return ret;
}

1503  Integer Inquiry

#include<stdio.h>
#include<string.h>
int main()
{
    int sum[102] = {0};
    char s[101];
    int i, j, len, c;
    gets_s(s);
    while (1)
    {
        if (s[0] == '0' && strlen(s) == 1)
            break;
        len = strlen(s);
        c = 0;
        for (i = len - 1, j = 0; i >= 0; i--, j++)
        {
            sum[j] += s[i]-'0' + c;
            c = sum[j] / 10;
            sum[j] = sum[j] % 10;
        }
        while (c > 0)
        {
            sum[j] += c;
            c = sum[j] / 10;
            sum[j] = sum[j] % 10;
            j++;
        }
        gets_s(s);
    }
    i = 101;
    while (sum[i] == 0)
        i--;
    while (i >= 0)
        printf("%d", sum[i--]);
    printf("\n");
    return 0;
}

2080  Calendar

#include<stdio.h>
int type(int);   /*判断是否为闰年*/
char week[7][10]={"Saturday","Sunday","Monday","Tuesday","Wednesday","Thursday","Friday"};
int year[2]={365,366};   /*year[0]表示非闰年的天数，year[1]表示闰年的天数。*/
int month[2][12]={31,28,31,30,31,30,31,31,30,31,30,31,
                  31,29,31,30,31,30,31,31,30,31,30,31};
/*month[0]表示非闰年里每个月的天数，month[1]表示闰年里每个月的天数。*/
int main()
{    int days,dayofweek;  /*days表示输入的天数，dayofweek表示星期几。*/
    int i,j;
    while(scanf("%d",&days)&&days!=-1)
    {    dayofweek=days%7;
        for(i=2000;days>=year[type(i)];i++)
            days-=year[type(i)];
        for(j=0;days>=month[type(i)][j];j++)
            days-=month[type(i)][j];
        printf("%d-%02d-%02d %s\n",i,j+1,days+1,week[dayofweek]);
    }
    return 0;
}
int type(int m)
{    if(m%4!=0||(m%100==0&&m%400!=0))
        return 0;
    else return 1;
}

2159  Ancient Cipher

解法1： //Time  16MS

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int main()
{    
    int i;
    int x[26], y[26];
    memset(x, 0, sizeof(x));
    memset(y, 0, sizeof(y));
    string input;
    cin >> input;
    for (i = 0; i < input.length(); i++)
    {
        x[input[i] - 'A']++;
    }
    cin >> input;
    for (i = 0; i < input.length(); i++)
    {
        y[input[i] - 'A']++;
    }
    sort(x, x + 26);
    sort(y, y + 26);
    for (i = 0; i < 26; i++)
        if (x[i] != y[i])
        {
            cout << "NO" << endl;
            return 0;
        }
    cout << "YES" << endl;
    return 0;
}

解法2： //Time   0MS

#include<stdio.h>
#include<stdlib.h>
int cmp(const void *a, const void *b)
{
    return *(int *)a > *(int *)b?1:-1;
}
int main()
{
    int i;
    int x[26], y[26];
    char input[101];
    for (i = 0; i < 26; i++)
        x[i] = y[i] = 0;
    gets_s(input);
    for (i = 0; input[i]; i++)
        x[input[i] - 'A']++;
    gets_s(input);
    for (i = 0; input[i]; i++)
        y[input[i] - 'A']++;
    qsort(x, 26, sizeof(x[0]), cmp);
    qsort(y, 26, sizeof(y[0]), cmp);
    for (i = 0; i < 26;i++)
        if (x[i] != y[i])
        {
            printf("NO\n");
            return 0;
        }
    printf("YES\n");
    return 0;
}

3069  Saruman's Army（贪心）

#include<stdio.h>
#include<algorithm>
using namespace std;
int x[1001],N,R;
void solve()
{
    for (int i = 0; i < N; i++)
        scanf_s("%d",&x[i]);
    sort(x, x + N);
    int i = 0,ans=0;
    while (i < N)
    {
        //s是没有被覆盖的最左的点的位置
        int s = x[i++];
        //一直向右前进直到距s的距离大于R的点
        while (i < N && x[i] <= s + R)
            i++;
        //p是新加上标记的点的位置
        int p = x[i - 1];
        //一直向右前进直到距p的距离大于R的点
        while (i < N && x[i] <= p + R)
            i++;
        ans++;

    }
    printf("%d\n",ans);
}
int main()
{
    while (scanf_s("%d%d", &R, &N) != EOF)
    {
        if (N != -1 && R != -1)
            solve();
        else
            break;
    }
    return 0;
}