Problem:
Score : 500 points

Problem Statement

Ibis is fighting with a monster.

The health of the monster is H.

Ibis can cast N kinds of spells. Casting the i-th spell decreases the monster's health by A_i, at the cost of B_i Magic Points.

The same spell can be cast multiple times. There is no way other than spells to decrease the monster's health.

Ibis wins when the health of the monster becomes 0 or below.

Find the minimum total Magic Points that have to be consumed before winning.

Constraints

1 \leq H \leq 10^4

1 \leq N \leq 10^3

1 \leq A_i \leq 10^4

1 \leq B_i \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H N
A_1 B_1
:
A_N B_N

Output

Print the minimum total Magic Points that have to be consumed before winning.

Sample Input 1

9 3
8 3
4 2
2 1

Sample Output 1

4

First, let us cast the first spell to decrease the monster's health by 8, at the cost of 3 Magic Points. The monster's health is now 1.

Then, cast the third spell to decrease the monster's health by 2, at the cost of 1 Magic Point. The monster's health is now -1.

In this way, we can win at the total cost of 4 Magic Points.

Sample Input 2

100 6
1 1
2 3
3 9
4 27
5 81
6 243

Sample Output 2

100

It is optimal to cast the first spell 100 times.

Sample Input 3

9999 10
540 7550
691 9680
700 9790
510 7150
415 5818
551 7712
587 8227
619 8671
588 8228
176 2461

Sample Output 3

139815

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.stream.IntStream;
import java.util.stream.LongStream;
import java.util.stream.Stream;

public class Main {

    public static void main(String[] args) throws IOException {
        new Main().solve();
    }

    private void solve() throws IOException {
        try {
//            solveA();
//            solveB();
//            solveC();
//            solveD();
            solveE();
            //			 solveF();
        } finally {
            if (in != null) {
                in.close();
            }
            if (out != null) {
                out.flush();
                out.close();
            }
        }

    }

    private void solveA() {
        int h = nextInt();
        int a = nextInt();
        int monster = (h / a) + ((h % a != 0) ? 1 : 0);
        out.println(monster);
    }

    private void solveB() {
        int h = nextInt();
        int n = nextInt();
        int total = IntStream.range(0, n).map(i -> nextInt()).sum();

        out.println(h <= total ? "Yes" : "No");
    }

    private void solveC() {
        int n = nextInt();
        int k = nextInt();
        long[] h = LongStream.range(0, n).map(i -> nextLong()).sorted().toArray();
        long res = 0;
        for (int i = 0; i < n - k; i++) {
            res += h[i];
        }
        out.println(res);
    }

    private void solveD() {
        long h = nextLong();
        int cnt = 0;
        while (h != 0) {
            h /= 2;
            cnt++;
        }
        long res = 1;
        for (int i = 1; i <= cnt; i++) {
            res <<= 1;
        }
        out.println(res - 1);
    }

    private void solveE() {
        int h = nextInt();
        int n = nextInt();
        int[][] wk = Stream.generate(() -> new int[]{nextInt(), nextInt()}).limit(n).toArray(int[][]::new);

        long[][] dp = new long[n + 2][h + 2];
        for (int i = 0; i <= n; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
            dp[i][0] = 0;
        }
        /*
        モンスターの体力が-0の時は魔法消費ゼロ
         */
        dp[0][0] = 0;
        for (int i = 0; i < n; i++) {
//        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= h; j++) {
                //ダメージ量（範囲を超えないように）
                /*
                1個前の魔法とwk[i-1][0]のダメージを与える魔法を使った場合(新規に魔法を使った)
                →ただし、魔法力0の場合があるの
                魔法を使わなかった場合
                wk[i-1]の魔法（同じ魔法）を使った場合(同じ魔法を繰り返した)
                 */
//                int from = Math.max(j - wk[i - 1][0], 0);
//                dp[i][j] = min(dp[i - 1][from] + wk[i - 1][1], dp[i - 1][j], dp[i][from] + wk[i - 1][1]);
                int from = Math.max(j - wk[i][0], 0);
                dp[i + 1][j] = min(dp[i][j], dp[i + 1][from] + wk[i][1]);
            }
        }
        out.println(dp[n][h]);
    }


    private void solveF() {
        int n = nextInt();

        out.println("");
    }

    /*
     * 階乗計算の最大値
     * もっと大きいのが必要ならここを修正
     */
    private final int FACTORIAL_NUM = 111111;
    /*
     * 階乗の事前計算（mod 済）
     */
    private long[] factorial = new long[FACTORIAL_NUM];
    /*
     * 逆数の事前計算（mod 済）
     */
    private long[] reverseFactorial = new long[FACTORIAL_NUM];
    private boolean isInitFactorial = false;

    /**
     * 階乗計算の事前累積和
     * [1, 1, 2, 3, 4, 5, … FACTORIAL_NUM]
     * mod済
     */
    private void initFactorial(long mod) {
        factorial[0] = 1;
        for (int i = 1; i < FACTORIAL_NUM; i++) {
            factorial[i] = (factorial[i - 1] * i) % mod;
        }
    }

    /**
     * 逆数の事前計算
     * もっと速い方法があったので修正
     */
    private void initReverseFactorial(long mod) {

        reverseFactorial[FACTORIAL_NUM - 1] = modInverse(factorial[FACTORIAL_NUM - 1], mod);
        for (int i = FACTORIAL_NUM - 2; i >= 0; i++) {
            /*
             * 以下を利用する。
             * これにより毎回modInverseしなくてよくなる
             * (N!)^(-1) * N -> ((N - 1)!)^(-1)
             * (N-1!)^(-1) * N -> ((N - 2)!)^(-1)
             */
            reverseFactorial[i] = (reverseFactorial[i + 1] * (i + 1)) % mod;
        }
        //		reverseFactorial[0] = 1;
        //		for (int i = 0; i < FACTORIAL_NUM; i++) {
        //			reverseFactorial[i] = modInverse(factorial[i], mod);
        //		}
    }

    /**
     * フェルマーの小定理を用いて逆元を求める。
     * ある数xのmod p（pは素数）の逆数x'はx' = x^(p-2)
     * 繰り返し二乗法を用いて計算する。
     * http://satanic0258.hatenablog.com/entry/2016/04/29/004730
     * {@link BigInteger#modInverse(BigInteger)}とどちらが速いか？
     *
     * @param x
     * @return
     */
    private long modInverse(long x, long mod) {
        long res = 1L;
        long k = mod - 2L;
        long y = x;
        while (k != 0) {
            if (k % 2 != 0) {
                res = (res * y) % mod;
            }
            y = (y * y) % mod;
            k /= 2;
        }
        return res;
    }

    private void initFactorialCalc(long mod) {
        if (isInitFactorial) {
            //初期化済みならreturn
            return;
        }
        //階乗の事前計算
        initFactorial(mod);
        //逆数の計算もしておく
        initReverseFactorial(mod);
        isInitFactorial = true;
    }

    /**
     * 組み合わせ
     * nCk -> n! / k!・(n-k)!
     *
     * @param n
     * @param k
     * @param mod
     * @return
     */
    private long combination(int n, int k, long mod) {
        initFactorialCalc(mod);
        if (k < 0 || n < k) {
            return 0;
        }

        //分子の計算
        long calResN = factorial[n];

        //分母の計算
        long calResNK = reverseFactorial[n - k];
        long calcResK = reverseFactorial[k];
        long denominator = (calResNK * calcResK) % mod;

        return (calResN * denominator) % mod;

    }

    /**
     * 順列
     * nPk -> n! / (n-k)!
     *
     * @param n
     * @param k
     * @return
     */
    private long permutation(int n, int k, long mod) {
        initFactorialCalc(mod);
        if (k < 0 || n < k) {
            return 0;
        }
        long calResN = factorial[n];
        long calResNK = reverseFactorial[n - k];
        return (calResN * calResNK) % mod;
    }

    /**
     * 多項係数
     * 文字aをp個,bをq個,cをr個, dをs個 あわせてn個を1列に並べるときの並べ方
     * n! / p!・q!・r!・s!
     *
     * @param n
     * @param strNum
     * @param mod
     * @return
     */
    private long multinominalCoefficient(int n, List<Integer> strNum, int mod) {
        initFactorialCalc(mod);

        long calResN = factorial[n];
        long calResDenomi = 1;
        for (Integer num : strNum) {
            calResDenomi *= reverseFactorial[num];
        }
        return (calResN * calResDenomi) % mod;
    }

    /**
     * 重複組合せ nHr （同次積）
     * nHr = (n+r-1)Cr
     * 異なるn個のものから重複を許してr個取る組合せの総数
     * 例：
     * リンゴ，ミカン，牛肉の3種類の果物があります．これらの中から6個の食材を買って帰ります．
     * このとき，何通りの買い方がありますか？ただし，含まれない食材があってもよいものとします
     *
     * @param n
     * @param r
     * @param mod
     * @return
     */
    private long homogeneousProduct(int n, int r, long mod) {
        if (n == 0 && r == 0) {
            return 1;
        }
        return combination(n + r - 1, r, mod);
    }

    private final PrintWriter out = new PrintWriter(System.out);
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        if (hasNextByte())
            return buffer[ptr++];
        else
            return -1;
    }

    private static boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr]))
            ptr++;
    }

    public boolean hasNext() {
        skipUnprintable();
        return hasNextByte();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public String next() {
        if (!hasNext())
            throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong() {
        if (!hasNext())
            throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }

    public long min(long... v) {
        long min = Long.MAX_VALUE;
        for (long i : v) min = Math.min(min, i);
        return min;
    }

    public long max(long... v) {
        long max = Long.MIN_VALUE;
        for (long i : v) max = Math.max(max, i);
        return max;
    }
}

class Combination {
    static final int MOD = (int) (Math.pow(10, 9) + 7);
    long[] fact;
    long[] invFact;
    long[] inv;

    public Combination(int n) {
        fact = new long[n + 1];
        invFact = new long[n + 1];
        inv = new long[n + 1];
        fact[0] = 1;
        fact[1] = 1;
        invFact[0] = 1;
        invFact[1] = 1;
        inv[1] = 1;
        for (int i = 2; i < n + 1; i++) {
            fact[i] = fact[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            invFact[i] = invFact[i - 1] * inv[i] % MOD;
        }
    }

    public long combination(int n, int k) {
        if (n < k || n < 0 || k < 0) {
            return 0;
        }
        return fact[n] * (invFact[k] * invFact[n - k] % MOD) % MOD;
    }
}