hdu5419 Victor and Toys

设数列s，位置为i的元素被区间覆盖的数目定义为∑s[j], j ≤ i。

则给定区间[l, r], 更新数列s,只需：

++s[l]，--s[r + 1]。

http://acm.hdu.edu.cn/showproblem.php?pid=5419

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef __int64 LL;

const int maxn = 5e4 + 10;

int w[maxn], cnt[maxn];
LL A, B;
int n, m;

LL gcd(LL a, LL b){
    if(!b) return a;
    return gcd(b, a % b);
}

LL f[maxn];

void init(){
    for(int i = 0; i < 3; i++) f[i] = 0;
    f[3] = 1;
    for(int i = 3; i < maxn; i++) f[i + 1] = f[i] * (i + 1) / (i - 2);
}

int main(){
    int T;
    scanf("%d", &T);
    init();
    while(T--){
        scanf("%d%d", &n, &m);
        memset(cnt, 0, sizeof cnt);
        for(int i = 1; i <= n; i++) scanf("%d", &w[i]);
        for(int i = 1, p, q; i <= m; i++){
            scanf("%d%d", &p, &q);
            ++cnt[p], --cnt[q + 1];
        }
        for(int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
        B = f[m];
        A = 0;
        for(int i = 1; i <= n; i++)    A += (LL)w[i] * f[cnt[i]];
        if(!A || !B){
            printf("0
");
            continue;
        }
        LL D = gcd(A, B);
        A /= D, B /= D;
        if(B == 1) printf("%I64d
", A);
        else printf("%I64d/%I64d
", A, B);
    }
    return 0;
}