正题

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题意

经典线段树题，不过是单点修改。

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解题思路

直接线段树

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代码

#include
     
      using namespace std;struct treenode{    int l,r,cover;}tree[400000];int n,m,x,y;char c;void build(int k,int a,int b)//线段树{    tree[k].l=a;tree[k].r=b;    if (a==b) return;    int wz=(a+b)/2;    build(k*2,a,wz);    build(k*2+1,wz+1,b);}int find(int k,int a,int b)//查找{    if (tree[k].r==b && tree[k].l==a)       return tree[k].cover;//返回值    if (b<=tree[k*2].r) return find(k*2,a,b);    else if (a>=tree[k*2+1].l) return find(k*2+1,a,b);    else return find(k*2,a,tree[k*2].r)+find(k*2+1,tree[k*2+1].l,b);}void updata(int k,int a,int num)//修改{    if (tree[k].r==a && tree[k].l==a) {        tree[k].cover+=num;//修改        return;    }    if (a<=tree[k*2].r) updata(k*2,a,num);    else updata(k*2+1,a,num);    tree[k].cover=tree[k*2].cover+tree[k*2+1].cover;//更新}int main(){    scanf("%d%d",&n,&m);    build(1,1,n);    for (int i=1;i<=m;i++)    {        scanf("\n%c%d%d",&c,&x,&y);        if (c=='M')        {            updata(1,x,y);        }        else        {            printf("%d\n",find(1,x,y));        }    }}