ZOJ Monthly, November 2012 G Gao The Sequence

Gao The Sequence

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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You are given a sequence of integers, A1,A2,...,An. And you are allowed a manipulation on the sequence to transform the origin sequence into another sequenceB1,B2,...,Bn(Maybe the two sequences are same ). The manipulation is specified as the following three steps:

1.Select an integer Ai and choose an arbitrary positive integer delta as you like.

2.Select some integers Aj satisfying j < i, let's suppose the selected integers are Ak1,Ak2,...,Akt , then subtract an arbitrary positive integer Di from Aki (1 ≤ i ≤ t) as long as sum(Di) = delta.

3.Subtract delta from Ai.

The manipulation can be performed any times. Can you find a way to transform A1,A2,...,An to B1,B2,...,Bn ?

Input

The input consist of multiple cases. Cases are about 100 or so. For each case, the first line contains an integer N(1 ≤ N ≤ 10000) indicating the number of the sequence. Then followed by N lines, ith line contains two integers Ai and Bi (0 ≤ Bi ≤ Ai ≤ 4294967296).

Output

Output a single line per case. Print "YES" if there is a certain way to transform Sequence A into Sequence B. Print "NO" if not.

Sample Input

3
3 2
4 2
5 2
3
2 0
7 1
3 1

Sample Output

YES
NO

分析：将原题转化为直接在差数组上操作。利用贪心的思路，若目前节余的值t+s[i-1]仍然<c[i]，则NO。一直没有NO就是YES。

#include<cstdio>
#include<cstring>
typedef long long u64;
u64 a[10010], b[10010], c[10010], s[10010], t;
int n;

int main() {
    int i;
    while (scanf("%d", &n) != EOF) {
        s[0] = 0;
        t = 0;
        for (i = 1; i <= n; ++i) {
            scanf("%lld%lld", &a[i], &b[i]);
            c[i] = a[i] - b[i];
            s[i] = s[i - 1] + c[i];
        }
        if (s[n]&1) {
            printf("NO\n");
            continue;
        }
        for (i = n; i >= 1; --i) {
            if (s[i - 1] + t < c[i]) {
                printf("NO\n");
                goto L;
            }
            if (s[i - 1] - c[i] <= t) {
                break;
            }
            t += c[i];
        }
        printf("YES\n");
L:
        ;
    }
    return 0;
}