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POJ1008 1013 1207 2105 2499（全部水题)

分类: IT文章 • 2022-07-17 15:09:29

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

char Tzo[][10] = {"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[20];
    int T;

    scanf("%d", &T);

    printf("%d
", T);
    for(int k=0; k<T; k++) {
        scanf("%d.%s%d", &day, szmonth, &year);

        for(int i=0; i<19; i++) if(strcmp(Haab[i], szmonth) == 0) month = i;

        cnt = year*365+month*20+day+1;

        int ty=0, td=0, tm=0;

        ty=(cnt-1+260)/260-1;
        tm=(cnt-1+20)%20;
        td=(cnt-1+13)%13+1;

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

    return 0;
}

View Code

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[5][20], _right[5][20], cond[5][20];

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 = 0;

    for(int i=0; i<len; i++) {
        if(s1[i] == 'A'+k) lef += j;
        if(s2[i] == 'A'+k) rig += j;
    }

    if(lef < rig) return 1;
    else if(lef == rig) return 0;
    else return -1;
}

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

    while(T--) {
        for(i=0; i<3; i++) {
            scanf("%s %s %s", _left[i], _right[i], cond[i]);
        }

        for(i=0; i<12; i++) {   //分别枚举12个字母
            for(j=-1; j<=1; j++) {  //轻重两种情况
                if(j==0) continue;

                for(c=0; c<3; c++) {    //三组
                    int res = cmp(_left[c], _right[c], i, j);
                    if( ! ((res == 1 && cond[c][0] == 'u') ||
                       (res == 0 && cond[c][0] == 'e') ||
                       (res == -1 && cond[c][0] == 'd')))
                       break;
                }
                if(c >= 3) break;
            }
            if(c >= 3) break;
        }

        printf("%c is the counterfeit coin and it is ", i+'A');
        if(j == -1) printf("heavy.
");
        else printf("light.
");
    }

    return 0;
}

View Code

POJ1207 The 3n + 1 problem

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

using namespace std;

const int maxn =  100000+10;

int hash[maxn];

int get(int n) {
    if(!hash[n]) {
        int t = n;
        int cnt = 0;
        while(true) {
            cnt++;
            if(n == 1) return (hash[t] = cnt);
            if(n % 2 == 1) n = 3*n+1;
            else n = n/2;
        }
    }
    else return hash[n];
}

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

    memset(hash, 0, sizeof(hash));

    while(scanf("%d %d", &n, &m) == 2) {
        a = n, b = m;
        if(a > b) swap(a, b);
        int max_v = -1;
        for(int i=a; i<=b; i++) {
            if(max_v < get(i)) {
                max_v = get(i);
            }
        }

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

    return 0;
}

View Code

POJ2105 IP Address

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

using namespace std;

const int maxn = 55;

char s[maxn];

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

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


        for(int i=0; i<4; i++) {
            int ans = 0;
            for(int j=0; j<8; j++) {
                ans += (s[i*8+j]-'0')*(1<<(7-j));
            }
            if(i == 0) printf("%d", ans);
            else printf(".%d", ans);
        }
        putchar('
');
    }

    return 0;
}

View Code

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=1; kase<=n; kase++) {
        cnt1 = cnt2 = 0;
        scanf("%d%d", &a, &b);

        while(a != b) {
            if(a > b) {
                cnt1 += a/b;
                a %= b;
                if(a == 0) {
                    cnt1--; break;
                }
            }
            else {
                cnt2 += b/a;
                b %= a;
                if(b == 0) {
                    cnt2--; break;
                }
            }
        }

        printf("Scenario #%d:
", kase);
        printf("%d %d

", cnt1, cnt2);
    }
    return 0;
}

View Code

POJ1008 1013 1207 2105 2499（全部水题)

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