LA2191:Potentiometers(树状数组)

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题意：

先给出一个数组，然后有两个操作

S x y 把第x个数改成y

M x y计算x~y个数的和

思路：

普通的树状数组，但不知道为什么会莫名其妙的一直WA

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;#define LS 2*i#define RS 2*i+1#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 5000005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit(x) (x&-x)LL c[N];int a[N],n;LL sum(int x){    LL ret = 0;    while(x>0)    {        ret+=c[x];        x-=lowbit(x);    }    return ret;}void add(int x,int d){    while(x<=n)    {        c[x]+=d;        x+=lowbit(x);    }}int main(){    int i,j,k,x,y,cas = 1;    char str[10];    scanf("%d",&n);    while(n)    {        printf("Case %d:\n",cas++);        for(i = 1; i<=n; i++)        {            scanf("%d",&a[i]);        }        MEM(c,0);        for(i = 1;i<=n;i++)        {            add(i,a[i]);        }        while(1)        {            scanf("%s",str);            if(!strcmp(str,"END")) break;            if(str[0]=='M')            {                scanf("%d%d",&x,&y);                LL ans = sum(y)-sum(x-1);                printf("%lld\n",ans);            }            else if(str[0]=='S')            {                scanf("%d%d",&x,&y);                add(x,y-a[x]);                a[x] = y;            }        }        scanf("%d",&n);        if(n)        printf("\n");    }    return 0;}

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