今天在coreforces上做的一题

4 4
1 2 3 4
1 1 4
2 1 4
1 2 4
2 1 3

17
12

大意：用fib去加到每个节点，求区域的和

代码如下：

#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;const int MAXNODE = 2097152; // == (1 << 21) - 1  depth:21
const int MAXS = 300003;
struct SOILD{int left,right; // 数组模拟满二叉树, 区间 [left,right]long value;
}Soldier[MAXNODE];
long fib[300005];
int father[MAXS]; // 存每一个士兵的父结点void BuildTree(int i,int left,int right,int devalue){ // 初始值为0Soldier[i].left = left;Soldier[i].right = right;Soldier[i].value = devalue;if (left == right){father[left] = i; // 可以瞬间知道第i个士兵的父节点是哪个return;}BuildTree(i*2, left, (int)floor((right+left)/2.0), 0);BuildTree(i*2+1, (int)floor((right+left)/2.0)+1, right, 0);
}void UpdataTree(int ri,int iKilln){ // 单点更新: 点p[i,i] 更新为iKilln (初始为0)if (ri == 1){Soldier[ri].value += iKilln;return;}Soldier[ri].value += iKilln; // iKilln 设为全局变量应该快一点UpdataTree(ri/2,iKilln);
}long res;
void Query(int i,int a,int b){ // [a,b]区间的和if (a == Soldier[i].left && b == Soldier[i].right){res += Soldier[i].value;return ;}if (a <= Soldier[i*2].right){ // 左端点小于右端点（几何含义：区间横跨过中点） 则区间需要分割if (b <= Soldier[i*2].right){Query(i*2, a, b); // 左边全包含}else{Query(i*2, a, Soldier[i*2].right); // 横跨取左}}if (b >= Soldier[i*2+1].left){if (a >= Soldier[i*2+1].left){Query(i*2+1, a, b); // 右边全包含}else{Query(i*2+1, Soldier[i*2+1].left, b); // 横跨取右}}
}int main(){int n_s,n_q;long iKilln,a,b;cin >> n_s >> n_q;BuildTree(1,1,n_s,0);fib[1]=1;fib[2]=1;for(int k2=3;k2<300005;k2++){fib[k2]=fib[k2-1]+fib[k2-2];}for (int ii = 1 ;ii <= n_s; ii++){ // 建树scanf("%d",&iKilln);  // iKilln 为第 i 个增加的杀敌数UpdataTree(father[ii],iKilln); // 瞬间找到士兵的结点序号}int order;for (int i = 1; i <= n_q; i++){scanf("%d%d%d",&order,&a,&b); // [a,b] ; a + kill b  if (order==2){res = 0;Query(1,a,b);printf("%d\n", res);}else{for(int kk=a;kk<=b;kk++)UpdataTree(father[kk],fib[kk-a+1]);}//for(int ttt=1;ttt<=n_s;ttt++)//  printf("%d;;",Soldier[father[ttt]].value);//  printf("\n");}return 0;
}

要取 long long/__int64；