• POJ 3468 A Simple Problem with Integers


    题意:一组数,两个操作,Q为查询l到r区间的和,C为更改l到r区间的值为原来的值加上c。

    解法:可以用线段树或树状数组。

    线段数成段更新,套模板改了改,万万没想到题目说int不会爆是骗我的……………… 

    线段树代码:

    #include<stdio.h>
    #include<iostream>
    #include<algorithm>
    #include<string>
    #include<string.h>
    #include<math.h>
    #include<limits.h>
    #include<time.h>
    #include<stdlib.h>
    #include<map>
    #include<queue>
    #include<set>
    #include<stack>
    #include<vector>
    #define LL long long
    #define lson l, m, rt << 1
    #define rson m + 1, r, rt << 1 | 1
    #define maxn 100005
    using namespace std;
    LL sum[maxn << 2], col[maxn << 2], point[maxn << 2];
    void pushdown(int rt)
    {
        if(col[rt])
        {
            col[rt << 1] += col[rt];
            col[rt << 1 | 1] += col[rt];
            sum[rt << 1] += col[rt] * point[rt << 1];
            sum[rt << 1 | 1] += col[rt] * point[rt << 1 | 1];
            col[rt] = 0;
        }
    }
    void pushup(int rt)
    {
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    }
    void build(int l, int r, int rt)
    {
        col[rt] = 0;
        if(l == r)
        {
            scanf("%I64d", &sum[rt]);
            point[rt] = 1;
            return ;
        }
        int m = (l + r) >> 1;
        build(lson);
        build(rson);
        point[rt] = point[rt << 1] + point[rt << 1 | 1];
        pushup(rt);
    }
    void update(int ll, int rr, int c, int l, int r, int rt)
    {
        if(ll <= l && rr >= r)
        {
            col[rt] += c;
            sum[rt] += (LL)c * point[rt];
            return ;
        }
        pushdown(rt);
        int m = (l + r) >> 1;
        if(ll <= m)
            update(ll, rr, c, lson);
        if(rr > m)
            update(ll, rr, c, rson);
        pushup(rt);
    }
    LL query(int ll, int rr, int l, int r, int rt)
    {
        if(ll <= l && rr >= r)
            return sum[rt];
        pushdown(rt);
        int m = (l + r) >> 1;
        LL ret = 0;
        if(ll <= m)
            ret += query(ll, rr, lson);
        if(rr > m)
            ret += query(ll, rr, rson);
        return ret;
    }
    int main()
    {
        int n, q;
        while(~scanf("%d%d", &n, &q))
        {
            build(1, n, 1);
            for(int i = 0; i < q; i++)
            {
                char ch[2];
                scanf("%s", ch);
                if(ch[0] == 'Q')
                {
                    int l, r;
                    scanf("%d%d", &l, &r);
                    printf("%I64d
    ", query(l, r, 1, n, 1));
                }
                else
                {
                    int l, r, c;
                    scanf("%d%d%d", &l, &r, &c);
                    update(l, r, c, 1, n, 1);
                }
            }
        }
        return 0;
    }
    

    树状数组解法:http://kenby.iteye.com/blog/962159

    写的很好……我就不多说了……

    树状数组代码:

    #include<stdio.h>
    #include<iostream>
    #include<algorithm>
    #include<string>
    #include<string.h>
    #include<math.h>
    #include<limits.h>
    #include<time.h>
    #include<stdlib.h>
    #include<map>
    #include<queue>
    #include<set>
    #include<stack>
    #include<vector>
    #define LL long long
    using namespace std;
    LL sum[100005], c1[100005], c2[100005];
    int n;
    inline int lowbit(int x)
    {
        return x & (-x);
    }
    void update(LL a[], int pos, LL val)
    {
        for(int i = pos; i <= n; i += lowbit(i))
            a[i] += val;
    }
    LL getsum(LL a[], int pos)
    {
        LL res = 0;
        for(int i = pos; i > 0; i -= lowbit(i))
            res += a[i];
        return res;
    }
    int main()
    {
        int m;
        while(~scanf("%d%d", &n, &m))
        {
            memset(c1, 0, sizeof(c1));
            memset(c2, 0, sizeof(c2));
            sum[0] = 0;
            for(int i = 1; i <= n; i++)
            {
                LL x;
                scanf("%lld", &x);
                sum[i] = sum[i - 1] + x;
            }
            for(int i = 0; i < m; i++)
            {
                char com[2];
                int l, r;
                scanf("%s%d%d", com, &l, &r);
                if(com[0] == 'Q')
                {
                    LL ans = sum[r] - sum[l - 1];
                    ans += (r + 1) * getsum(c1, r) - l * getsum(c1, l - 1);
                    ans -= getsum(c2, r) - getsum(c2, l - 1);
                    printf("%I64d
    ", ans);
                }
                else
                {
                    LL c;
                    scanf("%I64d", &c);
                    update(c1, l, c);
                    update(c1, r + 1, -c);
                    update(c2, l, c * l);
                    update(c2, r + 1, -c * (r + 1));
                }
            }
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Apro/p/4389361.html
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