10 | 10月 | 2015 | 超客的水墨烟雨

poj2393

maksyuki 发表于 oj 分类，标签: 基础算法-贪心

10 10月 2015

Yogurt factory

The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) cents to produce one unit of yogurt in week i. Yucky's factory, being well-designed, can produce arbitrarily many units of yogurt each week.

Yucky Yogurt owns a warehouse that can store unused yogurt at a constant fee of S (1 <= S <= 100) cents per unit of yogurt per week. Fortuitously, yogurt does not spoil. Yucky Yogurt's warehouse is enormous, so it can hold arbitrarily many units of yogurt.

Yucky wants to find a way to make weekly deliveries of Y_i (0 <= Y_i <= 10,000) units of yogurt to its clientele (Y_i is the delivery quantity in week i). Help Yucky minimize its costs over the entire N-week period. Yogurt produced in week i, as well as any yogurt already in storage, can be used to meet Yucky's demand for that week.

Input

* Line 1: Two space-separated integers, N and S.

* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.

Output

* Line 1: Line 1 contains a single integer: the minimum total cost to satisfy the yogurt schedule. Note that the total might be too large for a 32-bit integer.

Sample Input

4 5
88 200
89 400
97 300
91 500

Sample Output

126900

Hint

OUTPUT DETAILS:
In week 1, produce 200 units of yogurt and deliver all of it. In week 2, produce 700 units: deliver 400 units while storing 300 units. In week 3, deliver the 300 units that were stored. In week 4, produce and deliver 500 units.

Source

USACO 2005 March Gold

题目类型：贪心

算法分析：首先第一个用户需求一定要使用第一周生产的，之后的比较(aa[i-1] + s)和aa[i]的大小，选择其中较小的乘上第i周的用户需求即可

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 5e4 + 66;

LL aa[maxn], bb[maxn];

int main()
{
//    CFF;
    LL n, s;
    while (scanf("%lld%lld", &n, &s) != EOF)
    {
        for (LL i = 1; i <= n; i++)
            scanf ("%lld %lld", &aa[i], &bb[i]);

        LL res = aa[1] * bb[1];
        for (LL i = 2; i <= n; i++)
        {
            if (aa[i-1] + s < aa[i])
                res += (aa[i-1] + s) * bb[i];

            else res += aa[i] * bb[i];
        }

        printf ("%lld\n", res);
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 5e4 + 66;

LL aa[maxn], bb[maxn];

int main()

{

// CFF;

LL n, s;

while (scanf("%lld%lld", &n, &s) != EOF)

{

for (LL i = 1; i <= n; i++)

scanf ("%lld %lld", &aa[i], &bb[i]);

LL res = aa[1] * bb[1];

for (LL i = 2; i <= n; i++)

{

if (aa[i-1] + s < aa[i])

res += (aa[i-1] + s) * bb[i];

else res += aa[i] * bb[i];

}

printf ("%lld\n", res);

}

return 0;

}

Codeforces Round #324(Div.2) (2/5)

maksyuki 发表于比赛分类，标签: 比赛-codeforces

10 10月 2015

A Olesya and Rodion

Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.

Input

The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.

Output

Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.

Sample test(s)

input

3 2

output

712

题目类型：简单数学

算法分析：若t小于10，则一定存在满足条件的n位数，此时直接输出即可，若t等于10，且此时n为1，则输出-1，否则按照条件输出

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e6 +  6666;

int main()
{
//    CPPFF;
    int n, t;
    cin >> n >>t;
    if (t >= 2 && t <= 9)
    {
        for (int i = 1; i <= n; i++)
            cout << t;
        cout << endl;
    }

    else
    {
        if (n == 1)
            cout << "-1" << endl;

        else
        {
            for (int i = 1; i <= n / 2; i++)
                cout << t;

            if (n & 1)
                cout << "0";

            cout << endl;
        }
    }
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e6 + 6666;

int main()

{

// CPPFF;

int n, t;

cin >> n >>t;

if (t >= 2 && t <= 9)

{

for (int i = 1; i <= n; i++)

cout << t;

cout << endl;

}

else

{

if (n == 1)

cout << "-1" << endl;

else

{

for (int i = 1; i <= n / 2; i++)

cout << t;

if (n & 1)

cout << "0";

cout << endl;

}

return 0;

}

B Kolya and Tanya

Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.

More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place havea_i coins. If there is an integer i (0 ≤ i < n) such that a_i + a_i_+ n + a_i_+ 2n ≠ 6, then Tanya is satisfied.

Count the number of ways to choose a_i so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 10⁹ + 7. Two ways, a and b, are considered distinct if there is index i(0 ≤ i < 3n), such that a_i ≠ b_i (that is, some gnome got different number of coins in these two ways).

Input

A single line contains number n (1 ≤ n ≤ 10⁵) — the number of the gnomes divided by three.

Output

Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo10⁹ + 7.

Sample test(s)

input

output

input

output

680

Note

20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex):

题目类型：组合计数

算法分析：简单分析可以推出公式为(27 ^ n - 7 ^ n) % MOD，注意最后相减时要先加上一个MOD再取模，防止结果出现负数！！！

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define LL long long
#define ULL unsigned long long
#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1e4 +  6666;

LL mod_f (LL a, LL b, LL mod)
{
    LL res = 1;
    LL tt = (a + mod) % mod;
    while (b)
    {
        if (b & 1)
            res = res * tt % mod;

        b >>= 1;
        tt = tt * tt % mod;
    }
    return res;
}

int main()
{
//    CPPFF;
    LL n;
    cin >> n;
    LL resa = mod_f ((LL) 27, n, MOD);
    LL resb = mod_f ((LL) 7, n, MOD);
    cout << (resa - resb + MOD) % MOD << endl;
    return 0;
}

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define LL long long

#define ULL unsigned long long

#define DB(ccc) cout << #ccc << " = " << ccc << endl

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1e4 + 6666;

LL mod_f (LL a, LL b, LL mod)

{

LL res = 1;

LL tt = (a + mod) % mod;

while (b)

{

if (b & 1)

res = res * tt % mod;

b >>= 1;

tt = tt * tt % mod;

}

return res;

}

int main()

{

// CPPFF;

LL n;

cin >> n;

LL resa = mod_f ((LL) 27, n, MOD);

LL resb = mod_f ((LL) 7, n, MOD);

cout << (resa - resb + MOD) % MOD << endl;

return 0;

}

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超客的水墨烟雨

poj2393

Codeforces Round #324(Div.2) (2/5)

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