hdoj 1166 敌兵布阵

    暴力超时，这道题可以用线段树做，因为更新的是单个节点，我们也可以用数组数组来做，我将两种方法的代码都给出

    数组数组最适宜的用途就是区间求和和点的更新，但树状数组并不适用于区间的更新问题，也不是做不到，比较麻烦且难理解，有兴趣的可以看看这个http://blog.csdn.net/xindoo/article/details/8748410

//树状数组#include<stdio.h>int n,ans[50005],f[50005];int lowbit(int n){    return n&(-n);}void add(int i,int v){    while(i <= n)    {        ans[i] += v;        i += lowbit(i);    }}int query(int n){    int s = 0;    while(n)    {        s += ans[n];        n -= lowbit(n);    }    return s;}int main(){    int t;    scanf("%d",&t);    for(int j = 1;j <= t; j++)    {        scanf("%d",&n);        for(int l = 1; l <= n;l++)        {            scanf("%d",&f[l]);            f[l] += f[l-1];        }        for(int l = 1; l <= n; l++)            ans[l] = f[l]-f[l-lowbit(l)];        printf("Case %d:\n",j);        while(1)        {            char s[20];            int a,b;            scanf("%s",s);            if (s[0] == 'E') break;            scanf("%d%d",&a,&b);            if (s[0] == 'A') add(a,b);            if (s[0] == 'S') add(a,-b);            if (s[0] == 'Q') printf("%d\n",query(b)-query(a-1));        }    }    return 0;}

//线段树解法#include <stdio.h>#include <string.h>#define maxn 50005struct node{    int l, r, m;    int sum;}tree[maxn<<2];int a[maxn];void build(int l, int r, int o){    tree[o].l = l;    tree[o].r = r;    int m = (l+r)>>1;    tree[o].m = m;    if (l == r)    {        tree[o].sum = a[l];        return;    }    build(l, m, o<<1);    build(m+1, r, (o<<1)+1);    tree[o].sum = tree[o<<1].sum + tree[(o<<1)+1].sum; }void update(int l, int v, int o){    tree[o].sum += v;    if (tree[o].l == l && tree[o].r == l)        return;    if (l <= tree[o].m)        update(l, v, o<<1);    else        update(l, v, (o<<1)+1);}int query(int l, int r, int o){    if (tree[o].l == r && tree[o].r == r)        return tree[o].sum;    if (tree[o].m >= r)        return query(l, r, o<<1);    else if (tree[o].m <l)        return query(l, r, (o<<1)+1);    else        return query(l, tree[o].m, o<<1) + query(tree[o].m+1, r, (o<<1)+1);}int main(){    int t, n;    scanf("%d",&t);    for (int j = 1; j <= t; j++)    {        scanf("%d",&n);        for (int i = 1; i <= n; i++)            scanf("%d",&a[i]);        build(1, n, 1);        printf("Case %d:\n",j);        while(1)        {            char s[20];            int l, r;            scanf("%s",s);            if (s[0] == 'E')                break;            scanf("%d%d",&l,&r);            if (s[0] == 'A')                update(l, r, 1);            else if (s[0] == 'S')                update(l, -r, 1);            if (s[0] == 'Q')                printf("%d\n",query(l, r, 1));        }    }    return 0;}