POJ1008 1013 1207 2105 2499（全部水题)

做了一天水题，挑几个还算凑合的发上来。

POJ1008 Maya Calendar

分析：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

char Haab[][] = {"pop", "no", "zip", "zotz", "tzec", "xul", "yoxkin", "mol",
    "chen", "yax", "zac", "ceh", "mac", "kankin", "muan", "pax", "koyab", "cumhu", "uayet"};

char Tzo[][] = {"imix", "ik", "akbal", "kan", "chicchan", "cimi", "manik", "lamat",
    "muluk", "ok", "chuen", "eb", "ben", "ix", "mem", "cib", "caban", "eznab", "canac", "ahau"};

int main() {
    int year, day, cnt, month;
    char szmonth[];
    int T;

    scanf("%d", &T);

    printf("%d\n", T);
    for(int k=; k<T; k++) {
        scanf("%d.%s%d", &day, szmonth, &year);

        for(int i=; i<; i++) if(strcmp(Haab[i], szmonth) == ) month = i;

        cnt = year*+month*+day+;

        int ty=, td=, tm=;

        ty=(cnt-+)/-;
        tm=(cnt-+)%;
        td=(cnt-+)%+;

        printf("%d %s %d\n", td, Tzo[tm], ty);
    }

    return ;
}

POJ1013 Counterfeit Dollar

分析：

超级暴力：分别对A~L这12个字母枚举轻重两种情况。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

char _left[][], _right[][], cond[][];

int cmp(char s1[], char s2[], int k, int j) {
    /*
        * 用来比较天平的平衡
        * 右边up 返回 1
        * 平衡 返回 0
        * 右边down 返回 -1
    */
    int len = strlen(s1);
    int lef, rig;
    lef = rig = ;

    for(int i=; i<len; i++) {
        if(s1[i] == 'A'+k) lef += j;
        if(s2[i] == 'A'+k) rig += j;
    }

    if(lef < rig) return ;
    else if(lef == rig) return ;
    else return -;
}

int main(){
    int T, i, c, j;
    scanf("%d", &T);

    while(T--) {
        for(i=; i<; i++) {
            scanf("%s %s %s", _left[i], _right[i], cond[i]);
        }

        for(i=; i<; i++) {   //分别枚举12个字母
            for(j=-; j<=; j++) {  //轻重两种情况
                if(j==) continue;

                for(c=; c<; c++) {    //三组
                    int res = cmp(_left[c], _right[c], i, j);
                    if( ! ((res ==  && cond[c][] == 'u') ||
                       (res ==  && cond[c][] == 'e') ||
                       (res == - && cond[c][] == 'd')))
                       break;
                }
                if(c >= ) break;
            }
            if(c >= ) break;
        }

        printf("%c is the counterfeit coin and it is ", i+'A');
        if(j == -) printf("heavy.\n");
        else printf("light.\n");
    }

    return ;
}

POJ1207 The 3n + 1 problem

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

const int maxn =  +;

int hash[maxn];

int get(int n) {
    if(!hash[n]) {
        int t = n;
        int cnt = ;
        while(true) {
            cnt++;
            if(n == ) return (hash[t] = cnt);
            if(n %  == ) n = *n+;
            else n = n/;
        }
    }
    else return hash[n];
}

int main() {
    int a, b, n, m;

    memset(hash, , sizeof(hash));

    while(scanf("%d %d", &n, &m) == ) {
        a = n, b = m;
        if(a > b) swap(a, b);
        int max_v = -;
        for(int i=a; i<=b; i++) {
            if(max_v < get(i)) {
                max_v = get(i);
            }
        }

        printf("%d %d %d\n", n, m, max_v);
    }

    return ;
}

POJ2105 IP Address

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

const int maxn = ;

char s[maxn];

int main() {
    int n;
    scanf("%d", &n);

    while(n--) {
        scanf("%s", s);

        for(int i=; i<; i++) {
            int ans = ;
            for(int j=; j<; j++) {
                ans += (s[i*+j]-'')*(<<(-j));
            }
            if(i == ) printf("%d", ans);
            else printf(".%d", ans);
        }
        putchar('\n');
    }

    return ;
}

POJ2499 Binary Tree

分析：

用减的话会超时的。用除省时间。

#include <iostream>
#include <cstdio>

using namespace std;

int main(){
    int n, a, b;
    int cnt1, cnt2;
    scanf("%d", &n);
    for(int kase=; kase<=n; kase++) {
        cnt1 = cnt2 = ;
        scanf("%d%d", &a, &b);

        while(a != b) {
            if(a > b) {
                cnt1 += a/b;
                a %= b;
                if(a == ) {
                    cnt1--; break;
                }
            }
            else {
                cnt2 += b/a;
                b %= a;
                if(b == ) {
                    cnt2--; break;
                }
            }
        }

        printf("Scenario #%d:\n", kase);
        printf("%d %d\n\n", cnt1, cnt2);
    }
    return ;
}