Problem:
Score : 300 points

Problem Statement

There are N+1 towns. The i-th town is being attacked by A_i monsters.

We have N heroes. The i-th hero can defeat monsters attacking the i-th or (i+1)-th town, for a total of at most B_i monsters.

What is the maximum total number of monsters the heroes can cooperate to defeat?

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{N+1}
B_1 B_2 ... B_N

Output

Print the maximum total number of monsters the heroes can defeat.

Sample Input 1

2
3 5 2
4 5

Sample Output 1

9

If the heroes choose the monsters to defeat as follows, they can defeat nine monsters in total, which is the maximum result.

The first hero defeats two monsters attacking the first town and two monsters attacking the second town.

The second hero defeats three monsters attacking the second town and two monsters attacking the third town.

Sample Input 2

3
5 6 3 8
5 100 8

Sample Output 2

22

Sample Input 3

2
100 1 1
1 100

Sample Output 3

3

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


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;

public class Main {

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

	private void solve() throws IOException {
		try {
			//			solveA();
			//			solveB();
			//			solveB2();
			solveC();
			//			 solveD();
			//			 solveE();
			//			 solveF();
		} finally {
			if (in != null) {
				in.close();
			}
			if (out != null) {
				out.flush();
				out.close();
			}
		}

	}

	private void solveA() {
		long a = nextLong();
		long b = nextLong();
		if (Math.abs(a - b) == Math.abs(b - a) && Math.abs(a - b) % 2 == 0) {
			long wk = Math.abs(a - b) / 2;
			//			if (a > wk)
			//				out.println(a - wk);
			//			else
			//				out.println(a + wk);
			long wk2 = Math.max(a, b) - wk;
			out.println(wk2);
			//			out.println((a + b) / 2);
		} else {
			out.println("IMPOSSIBLE");
		}
	}

	private void solveB() {
		int n = nextInt();
		int[] wk = IntStream.range(0, n).map(i -> nextInt()).toArray();
		int cnt = 0;
		for (int i = 0; i < wk.length; i++) {
			if (wk[i] != i + 1)
				cnt++;
		}
		if (cnt <= 2)
			out.println("YES");
		else
			out.println("NO");
	}

	private void solveB2() {
		int n = nextInt();
		int[] wk = IntStream.range(0, n).map(i -> nextInt()).toArray();
		boolean d = false;
		int b = 0;
		int bI = 0;
		int sI = 0;
		for (int i = 1; i < wk.length; i++) {
			if (!d && wk[i - 1] > wk[i]) {
				d = true;
				b = wk[i];
				bI = i;
				sI = i - 1;
			}
			if (d && b > wk[i]) {
				b = wk[i];
				bI = i;
			}
		}
		int wkT = wk[sI];
		wk[sI] = wk[bI];
		wk[bI] = wkT;

		for (int i = 1; i < wk.length; i++) {
			if (wk[i - 1] > wk[i]) {
				out.println("NO");
				return;
			}
		}
		out.println("YES");
	}

	private void solveC() {
		int n = nextInt();
		int[] a = IntStream.range(0, n + 1).map(i -> nextInt()).toArray();
		int[] b = IntStream.range(0, n).map(i -> nextInt()).toArray();

		int total = Arrays.stream(a).sum();
		for (int i = 0; i < n; i++) {
			if (b[i] >= a[i]) {
				b[i] -= a[i];
				a[i] = 0;
			} else {
				a[i] -= b[i];
				b[i] = 0;
			}
			if (b[i] > 0) {
				if (a[i + 1] >= b[i]) {
					a[i + 1] -= b[i];
				} else {
					a[i + 1] = 0;
				}
			}
		}
		int fix = Arrays.stream(a).sum();
		out.println(total - fix);
	}

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

		out.println("");
	}

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

		out.println("");
	}

	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();
		}
	}
}