训练指南第二章-基础问题

训练指南第二章-基础问题 P170

	2 / 4	Problem A	UVA 10943	How do you add?
	1 / 2	Problem B	UVA 10780	Again Prime? No Time.
	1 / 1	Problem C	UVA 10892	LCM Cardinality
	1 / 4	Problem D	UVA 11752	The Super Powers
	1 / 3	Problem E	UVA 11076	Add Again
	1 / 1	Problem F	UVA 11609	Teams
	1 / 3	Problem G	POJ 2402	Palindrome Numbers(LA2889)
	1 / 1	Problem H	UVA 11489	Integer Game
	1 / 6	Problem I	UVA 10791	Minimum Sum LCM
	1 / 1	Problem J	UVA 11461	Square Numbers
	1 / 2	Problem K	UVA 11388	GCD LCM
	1 / 2	Problem L	UVA 11889	Benefit
	1 / 1	Problem M	UVA 1319	Maximum
	1 / 1	Problem N	UVA 1315	Crazy tea party

 

 

 

 

 

 

 

 

 

 

 

 

 

 

UVA 11388 GCD LCM

显然在合理的情况下，a=G，b=L

#include <bits/stdc++.h>

int main() {
    int T;
    scanf ("%d", &T);
    while (T--) {
        int g, l;
        scanf ("%d%d", &g, &l);
        long long f = 1ll * g * l;
        if (l < g || l % g != 0) {
            puts ("-1");
        } else {
            printf ("%d %d
", g, l);
        }
    }
    return 0;
}

UVA 11889 Benefit

解法一：分解质因数，因为,,，所以在ai小于最大值时bi为最大值，否则为0.

解法二：数学方法，tb=C/A，当g=GCD(A, tb) != 1，a /= g. tb *= .（不是很懂）

#include <bits/stdc++.h>

void get_prime(int n, std::map<int, int> &mp) {
    int m = (int) sqrt (n + 0.5);
    for (int i=2; i<=m; ++i) {
        if (n % i == 0) {
            int cnt = 0;
            while (n % i == 0) {
                n /= i;
                mp[i]++;
            }
        }
    }
    if (n != 1) {
        mp[n]++;
    }
}

int _pow(int x, int n) {
    int ret = 1;
    while (n) {
        if (n & 1) {
            ret = ret * x;
        }
        n >>= 1;
        x = x * x;
    }
    return ret;
}

int solve(int a, int c) {
    std::map<int, int> mpa, mpc;
    get_prime (a, mpa);
    get_prime (c, mpc);
    int na = mpa.size (), nc = mpc.size ();
    if (na > nc || c % a != 0) {
        return -1;
    } else {
        int ret = 1;
        int m = (int) sqrt (c + 0.5);
        std::map<int, int>::iterator it;
        for (it=mpc.begin (); it!=mpc.end (); ++it) {
            if (!mpa.count (it->first) || mpa[it->first] < it->second) {
                ret *= _pow (it->first, it->second);
            }
        }
        return ret;
    }
}

int main() {
    int T;
    scanf ("%d", &T);
    while (T--) {
        int a, c;
        scanf ("%d%d", &a, &c);
        int ans = solve (a, c);
        if (ans != -1) {
            printf ("%d
", ans);
        } else {
            puts ("NO SOLUTION");
        }
    }
    return 0;
}

#include <bits/stdc++.h>

int GCD(int a, int b) {
    return b ? GCD (b, a % b) : a;
}

int solve(int a, int c) {
    if (a > c || c % a != 0) {
        return -1;
    }
    int tb = c / a;
    int g = GCD (a, tb);
    int mul = 1;
    while (g != 1) {
        mul *= g;
        a /= g;
        g = GCD (a, tb);
    }
    return tb * mul;
}

int main() {
    int T;
    scanf ("%d", &T);
    while (T--) {
        int a, c;
        scanf ("%d%d", &a, &c);
        int ans = solve (a, c);
        if (ans != -1) {
            printf ("%d
", ans);
        } else {
            puts ("NO SOLUTION");
        }
    }
    return 0;
}

UVA 10943 How do you add?

这个用隔板法，

#include <bits/stdc++.h>

int binom[205][205];
const int MOD = 1000000;

void init_binom() {
    for (int i=1; i<=200; ++i) {
        binom[i][0] = binom[i][i] = 1;
        for (int j=1; j<i; ++j) {
            binom[i][j] = binom[i-1][j-1] + binom[i-1][j];
            if (binom[i][j] >= MOD) {
                binom[i][j] -= MOD;
            }
        }
    }
}

int main() {
    init_binom ();
    int n, k;
    while (scanf ("%d%d", &n, &k) == 2) {
        if (!n && !k) {
            break;
        }
        printf ("%d
", binom[n+k-1][k-1]);
    }
    return 0;
}

UVA 10780 Again Prime? No Time.

分解质因数，取n!与m的某个质因数的个数商最小值。n!的某个质因数的个数=

#include <bits/stdc++.h>

int main() {
    int T;
    scanf ("%d", &T);
    for (int cas=1; cas<=T; ++cas) {
        int m, n;
        scanf ("%d%d", &m, &n);
        int ans = 0x3f3f3f3f;
        int i = 2;
        while (m != 1) {
            int cnt = 0;
            while (m % i == 0) {
                cnt++;
                m /= i;
            }
            if (cnt) {
                int nn = n;
                int cnt2 = 0;
                while (nn) {
                    cnt2 += nn / i;
                    nn /= i;
                }
                ans = std::min (ans, cnt2 / cnt);
            }
            i++;
        }
        printf ("Case %d:
", cas);
        if (ans) {
            printf ("%d
", ans);
        } else {
            puts ("Impossible to divide");
        }
    }
    return 0;
}

UVA 10892 LCM Cardinality

分解质因数，暴力枚举~

#include <bits/stdc++.h>

typedef long long ll;

int GCD(int a, int b) {
    return b ? GCD (b, a % b) : a;
}

ll LCM(int a, int b) {
    return 1ll * a / GCD (a, b) * b;
}

int main() {
    int n;
    while (scanf ("%d", &n) == 1) {
        if (!n) {
            break;
        }
        std::vector<int> vec;
        int m = (int) sqrt (n + 0.5);
        for (int i=1; i<=m; ++i) {
            if (n % i == 0) {
                if (i != n / i) {
                    vec.push_back (i);
                    vec.push_back (n / i);
                } else {
                    vec.push_back (i);
                }
            }
        }
        int ans = 1; //n n
        for (int i=0; i<vec.size (); ++i) {
            for (int j=i+1; j<vec.size (); ++j) {
                if (LCM (vec[i], vec[j]) == n) {
                    ans++;
                }
            }
        }
        printf ("%d %d
", n, ans);
    }
    return 0;
}

UVA 11752 The Super Powers

只要指数不是素数才可以符合条件，技巧：取对数来求最大的指数

#include <bits/stdc++.h>

typedef unsigned long long ull;
int nprime[105];

bool is_prime(int n) {
    int m = (int) sqrt (n + 0.5);
    for (int i=2; i<=m; ++i) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}

ull _pow(ull x, int n) {
    ull ret = 1;
    while (n) {
        if (n & 1) {
            ret = ret * x;
        }
        n >>= 1;
        x *= x;
    }
    return ret;
}

int main() {
    int tot = 0;
    for (int i=4; i<=64; ++i) {
        if (!is_prime (i)) {
            nprime[tot++] = i;
        }
    }
    std::map<ull, bool> mp;
    int bound = (1 << 16);
    for (int i=2; i<=bound; ++i) {
        double maxn = log (pow (2.0, 64.0)-1) / log (i);
        for (int j=0; j<tot; ++j) {
            int n = nprime[j];
            if (n >= maxn) {
                break;
            }
            mp[_pow ((ull) i, n)] = true;
        }
    }
    mp[1] = true;
    std::map<ull, bool>::iterator it;
    for (it=mp.begin (); it!=mp.end (); ++it) {
        printf ("%llu
", it->first);
    }
    return 0;
}

UVA 11076 Add Again

设有重复元素的全排列数为x，，所以，不考虑位置的话，总和为,然后考虑进位累加的问题。

#include <bits/stdc++.h>

typedef unsigned long long ull;

int main() {
    int a[13], cnt[15];
    ull f[13];
    f[0] = f[1] = 1;
    for (int i=2; i<=12; ++i) {
        f[i] = f[i-1] * i;
    }
    int n;
    while (scanf ("%d", &n) == 1) {
        if (!n) {
            break;
        }
        memset (cnt, 0, sizeof (cnt));
        int m = 0;
        for (int i=1; i<=n; ++i) {
            int x;
            scanf ("%d", &x);
            if (!cnt[x]) {
                a[m++] = x;
            }
            cnt[x]++;
        }
        ull sum = 0;
        for (int i=0; i<m; ++i) {
            ull div = f[cnt[a[i]]-1];
            for (int j=0; j<m; ++j) {
                if (i != j) {
                    div *= f[cnt[a[j]]];
                }
            }
            sum += f[n-1] / div * a[i];
        }
        //printf ("$%llu
", sum);
        ull ans = 0;
        for (int i=1; i<=n; ++i) {
            ans += sum;
            sum *= 10;
        }
        printf ("%llu
", ans);
        /*
        ull tmp = 0;
        std::vector<ull> ans;
        for (int i=1; i<=n; ++i) {
            ans.push_back ((sum + tmp) % 10);
            tmp = (sum + tmp) / 10;
        }
        if (tmp) {
            printf ("%llu", tmp);
        }
        for (int i=ans.size ()-1; i>=0; --i) {
            printf ("%llu", ans[i]);
        }
        puts ("");
        */
    }
    return 0;
}

UVA 11609 Teams

#include <bits/stdc++.h>

const int MOD = 1e9 + 7;

int pow_mod(int x, int n) {
    int ret = 1;
    while (n) {
        if (n & 1) {
            ret = 1ll * ret * x % MOD;
        }
        n >>= 1;
        x = 1ll * x * x % MOD;
    }
    return ret;
}

int main() {
    int T;
    scanf ("%d", &T);
    for (int cas=1; cas<=T; ++cas) {
        int n;
        scanf ("%d", &n);
        printf ("Case #%d: %d
", cas, 1ll * pow_mod (2, n - 1) * n % MOD);
    }
    return 0;
}

POJ 2402 Palindrome Numbers(LA2889)

按照回文串长度分组，找到是第n/num[i]+1组的第n%num[i]-1个位置，因为每组都从1000...0001开始，很容易找到规律。

#include <bits/stdc++.h>

typedef long long ll;
ll num[32];

ll tenp(int m) {
    ll ret = 1;
    while (m--) {
        ret *= 10;
    }
    return ret;
}

int main() {
    ll s = 9;
    num[1] = num[2] = s;
    for (int i=3; i<31; i+=2) {
        s *= 10;
        num[i] = num[i+1] = s;
    }
    int n;
    while (scanf ("%d", &n) == 1) {
        if (!n) {
            break;
        }
        int i;
        for (i=1; i<32 && n > num[i]; ++i) {
            n -= num[i];
        }
        int l = i / 2 + (i & 1);
        ll x = tenp (l - 1) + n - 1;
        printf ("%I64d", x);
        if (i & 1) {
            x /= 10;
        }
        while (x) {
            printf ("%d", x % 10);
            x /= 10;
        }
        puts ("");
    }

    return 0;
}

UVA 11489 Integer Game

除了第一次特判外，接下来只能取3，6，9（保证剩余数字是3的倍数），判断奇偶就可以了。

#include <bits/stdc++.h>

const int N = 1e3 + 5;
char str[N];
int cnt[10];

bool judge() {
    memset (cnt, 0, sizeof (cnt));
    int sum = 0;
    for (int i=0; str[i]; ++i) {
        sum += str[i] - '0';
        cnt[str[i]-'0']++;
    }
    bool flag = false;
    int t = sum % 3;
    for (int i=t; i<10; i+=3) {
        if (cnt[i]) {
            cnt[i]--;
            flag = true;
            break;
        }
    }
    if (!flag) {
        return false;
    }
    int tot = cnt[3] + cnt[6] + cnt[9];
    return !(tot & 1);
}

int main() {
    int T;
    scanf ("%d", &T);
    for (int cas=1; cas<=T; ++cas) {
        scanf ("%s", str);
        printf ("Case %d: %c
", cas, judge () ? 'S' : 'T');
    }
    return 0;
}

UVA 10791 Minimum Sum LCM

这题题目说的是set集合，不仅仅只是两个数字，结论是所有数字两两互质时最小，即分解质因数后，求

#include <bits/stdc++.h>

long long solve(int n) {
    long long ret = 0;
    int cnt = 0;
    int m = (int) sqrt (n + 0.5);
    int nn = n;
    for (int i=2; i<=m; ++i) {
        if (n % i == 0) {
            cnt++;
            long long tmp = 1;
            while (n % i == 0) {
                n /= i;
                tmp *= i;
            }
            ret += tmp;
        }
    }
    if (!cnt || (cnt == 1 && n == 1)) {
        return 1ll + nn;
    } else {
        if (n != 1) {
            ret += n;
        }
        return ret;
    }
}

int main() {
    int n, cas = 0;
    while (scanf ("%d", &n) == 1) {
        if (!n) {
            break;
        }
        printf ("Case %d: %lld
", ++cas, solve (n));
    }
    return 0;
}

UVA 11461 Square Numbers

前缀思想

#include <bits/stdc++.h>

const int N = 1e5 + 5;
int sum[N];

bool ok(int n) {
    int m = (int) sqrt (n);
    return m * m == n;
}

void init() {
    sum[0] = 0;
    for (int i=1; i<=100000; ++i) {
        sum[i] = sum[i-1];
        if (ok (i)) {
            sum[i]++;
        }
    }
}

int main() {
    init ();
    int a, b;
    while (scanf ("%d%d", &a, &b) == 2) {
        if (!a && !b) {
            break;
        }
        printf ("%d
", sum[b] - sum[a-1]);
    }
    return 0;
}

UVA 1319 Maximum

都乘以。，。因为p是偶数，贪心思想一些数取-1，一些取a，还有一个在范围内。

#include <bits/stdc++.h>

typedef long long ll;

int main() {
    int m, p, a, b;
    while (scanf ("%d%d%d%d", &m, &p, &a, &b) == 4) {
        int sum = a * b;
        int x = 0, y = 0;
        for (int i=1; i<m; ++i) {
            if (sum >= a) {
                sum -= a;
                x++;
            } else {
                sum++;
                y++;
            }
        }
        double sa = sqrt (a * 1.0);
        double ans = y / pow (sa, p) + x * pow (sa, p);
        ans += pow (sum / sa, p);
        printf ("%d
", (int) (ans + 0.5));
    }
    return 0;
}

UVA 1315 Crazy tea party

找规律，靠近1的往1挪，靠近n的往n挪。

#include <bits/stdc++.h>

int main() {
    int T;
    scanf ("%d", &T);
    while (T--) {
        int n;
        scanf ("%d", &n);
        int m = (n - 1) / 2;
        int ans = (1 + m) * m;
        if (n & 1) {
            ans -= m;
        }
        printf ("%d
", ans);
    }
    return 0;
}