CodeForces Round #179 (295A)

题目链接：http://codeforces.com/problemset/problem/295/A

我的做法，两次线段树

#include <cstdio>
#include <cstring>

const int N = 100005;

long long sumv[N * 3];
long long add[N * 3];
long long a[N];

struct opera {
    int l, r;
    long long d;
} op[N];

void pushdown(int o, int l, int r)
{
    int m = (l + r) >> 1;
    sumv[o << 1] += add[o] * (m - l + 1);
    add[o << 1] += add[o];
    sumv[o << 1 | 1] += add[o] * (r - m);
    add[o << 1 | 1] += add[o];
    add[o] = 0;
}

int ql, qr;
long long val;
void update(int o, int l, int r)
{
    if (ql <= l && qr >= r) {
        sumv[o] += (r - l + 1) * val;
        add[o] += val;
        return ;
    }
    if (add[o]) pushdown(o, l, r);
    int m = (l + r) >> 1;
    if (m < qr) update(o << 1 | 1, m + 1, r);
    if (m >= ql) update(o << 1, l, m);
    sumv[o] = sumv[o << 1] + sumv[o << 1 | 1];
}

int q;
long long query(int o, int l, int r)
{
    if (l == r) return sumv[o];
    if (add[o]) pushdown(o, l, r);
    int m = (l + r) >> 1;
    if (m < q) return query(o << 1 | 1, m + 1, r);
    else return query(o << 1, l, m);
}

int main()
{
    //freopen("a.in", "r", stdin);

    int n, m, k;
    int i;
    scanf("%d%d%d", &n, &m, &k);
    for (i = 1; i <= n; ++i) scanf("%lld", a + i);
    for (i = 1; i <= m; ++i) {
        scanf("%d%d%lld", &op[i].l, &op[i].r, &op[i].d);
    }
    val = 1;
    for (i = 1; i <= k; ++i) {
        scanf("%d%d", &ql, &qr);
        update(1, 1, m);
    }
    for (i = 1; i <= m; ++i) {
        q = i;
        op[i].d *= query(1, 1, m);
    }
    memset(sumv, 0, sizeof sumv);
    memset(add, 0, sizeof add);
    for (i = 1; i <= m; ++i) {
        ql = op[i].l;
        qr = op[i].r;
        val = op[i].d;
        update(1, 1, n);
    }
    for (i = 1; i <= n; ++i) {
        q = i;
        printf("%lld", a[i] + query(1, 1, n));
        if (i != n) printf(" ");
    }
    return 0;
}

　　

后来看了学长的代码，又写了一遍.......：

#include <cstdio>

const int N = 100005;

long long l[N];
long long r[N];
long long d[N];
long long a[N];
long long op[N];
long long b[N];

main()
{
    //freopen("in.txt", "r", stdin);
    long long n, m, k;
    scanf("%lld%lld%lld", &n, &m, &k);
    for (int i = 1; i <= n; ++i)
        scanf("%lld", &a[i]);
    for (int i = 1; i <=  m; ++i)
        scanf("%lld%lld%lld", &l[i], &r[i], &d[i]);
    long long x, y;
    for (int i = 1; i <= k; ++i) {
        scanf("%lld%lld", &x, &y);
        ++op[x];
        --op[y+1];
    }
    for (int i = 1; i <= m; ++i) {
        op[i] += op[i - 1];
        d[i] *= op[i];
    }
    for (int i = 1; i <= m; ++i) {
        b[l[i]] += d[i];
        b[r[i]+1] -= d[i];
    }
    for (int i = 1; i <= n; ++i) {
        b[i] += b[i - 1];
        a[i] += b[i];
    }
    printf("%lld", a[1]);
    for (int i = 2; i <= n; ++i)
        printf(" %lld", a[i]);
}