UVALive 2191 Potentiometers （树状数组）

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=192

给你一个长度为n的数组A，有两种操作，操作一：(S x y) 将A[x]的大小置为y，操作二：(M x y) 求A[x], A[x+1] ... A[y］的和

用一个额外的数组tmp存储数组A，然后将 S x y转换为树状数组的 add x y-tmp[x]

#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 200010
using namespace std;

typedef long long LL;

LL c[maxn];
int n, arr[maxn];

int lowbit(int x);

void add(int x, int u);

LL sum(int x);

int main(void)
{
    int ca = 0;
    while (scanf("%d", &n), n)
    {
        memset( c, 0, sizeof(c));
        int x, y;
        char ope[10];
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", arr + i);
            add( i, arr[i]);
        }
        if (ca)
            putchar(10);
        printf("Case %d:
", ++ca);
        while (scanf("%s", ope), ope[0] != 'E')
        {
            scanf("%d %d", &x, &y);
            if (ope[0] == 'S')
            {
                add( x, y - arr[x]);
                arr[x] = y;
            }
            else
            {
                printf("%lld
", sum(y) - sum(x - 1));
            }
        }
    }
    return 0;
}

int lowbit(int x)
{
    return x&(-x);
}

LL sum(int x)
{
    LL result = 0;
    while (x > 0)
    {
        result += c[x];
        x -= lowbit(x);
    }
    return result;
}

void add(int x, int u)
{
    while (x <= n)
    {
        c[x] += u;
        x += lowbit(x);
    }
}