POJ3468 A Simple Problem with Integers (树状数组 | 线段树)

题目链接：http://poj.org/problem?id=3468

方法一：树状数组   区间修改区间查询问题。

讲一下关于树状数组区间修改和区间查询的问题的解决方法。

设a[i]为原数组，d[i] = a[i] - a[i-1],其中a[0] = 0;   

递推一下：    d[2] = a[2] - a[1]

                      d[3] = a[3] - a[2]

                      d[4] = a[4] - a[3]

                      .......

                      .......

                      ......

                      .......

                      d[x] = a[x] - a[x-1]

通过上面递推式可以发现，a[x] = d[x]+a[x-1],其中a[x-1] = d[x-1] + a[x-2]，，所以不难发现 a[x] = (d[1]+d[2]+d[3}+......+d[x])，即a[x] = segma(d[i]),其中1<=i<=x；

所以 segma(a[i])  (1<=i<=x)  = segma(segma(d[j], 1 <= j <= i));    观察式子可知，其中一共有x个d[1]相加，x-1个d[2]相加等等，由此可以推出规律：

segma(a[i]   1 <= i <= x) = segma((x-i+1) * d[i])；   即： segma(a[i]  1 <= i <= x) = (x+1) * segma(d[i]  1<= i <= x) - segma(d[i] * i    1 <= i <= x)；

其中d[i]数组存储的是每次的增量， 再设一个变量sum[i],表示前i个原数组a[]中的和，ans[i]表示前i项的总和。

则d[i]对ans[x]的贡献值就是  d[i] (x - i + 1).

AC代码如下：

#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#define inf 0x3f3f3f3f#define mod 10000000007 #define ll long long#define ld long double#define ls rt << 1#define rs rt << 1 | 1#define pi acos(-1.0)using namespace std;const int maxn = 1e5 + 5;ll c1[maxn],c2[maxn],a[maxn],sum[maxn];int n,Q;int lowbit(int k){return k & -k;}void update(ll *array, int k, int num){while(k <= n){array[k] += num;k += lowbit(k);}}ll query(ll *array, int k){ll ans = 0;while(k){ans += array[k];k -= lowbit(k);}return ans;}int main(){while(~scanf("%d%d",&n,&Q)){memset(c2,0,sizeof(c2));memset(c1,0,sizeof(c1)); memset(sum,0,sizeof(sum));for(int i = 1; i <= n; i ++){scanf("%lld",&a[i]);}for(int i = 1; i <= n; i ++){sum[i] = sum[i-1] + a[i];} char str[10];while(Q --){scanf("%s",str);int b,c,d;if(str[0] == 'Q'){scanf("%d%d",&b,&c);ll ans = sum[c] - sum[b-1];ans += (c+1) * query(c1,c) - query(c2,c);ans -= (b-1+1) * query(c1,b-1) - query(c2,b-1); printf("%lld\n",ans);}else{scanf("%d%d%d",&b,&c,&d);update(c1,b,d);update(c2,b,b*d);update(c1,c+1,-d); update(c2,c+1,(c+1) * (-d));}}}return 0;}

线段树代码：区间更新求值就行，注意下推标记更新子节点。

AC代码如下：

#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#define inf 0x3f3f3f3f#define ll long long#define ls rt << 1#define rs rt << 1 | 1#define pi acos(-1.0)using namespace std;const int maxn = 1e5 + 5;ll sum[maxn << 2],add[maxn << 2],d[maxn];int N,Q;void PushUp(int rt){sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void PushDown(int rt, int ln, int rn){if(add[rt]){add[ls] += add[rt];add[rs] += add[rt];sum[ls] += add[rt] * ln; sum[rs] += add[rt] * rn;add[rt] = 0;}}void Build(int l, int r, int rt){if(l == r){sum[rt] = d[l];return;}int m = (l + r) >> 1;   // printf("2333\n");Build(l,m,rt << 1);Build(m+1,r,rt << 1 | 1);PushUp(rt);}void Update(int L, int R, int C, int l, int r, int rt){if(L <= l && r <= R){sum[rt] += (r - l + 1) * C;add[rt] += C;return ;}int m = (l + r) >> 1;PushDown(rt,m-l+1,r-m);if(L <= m) Update(L,R,C,l,m,ls);if(R > m) Update(L,R,C,m+1,r,rs);PushUp(rt);}ll Query(int L, int R, int l, int r, int rt){if(L <= l && r <= R) return sum[rt];int m = (l + r) >> 1;PushDown(rt,m-l+1,r-m); ll ans = 0;if(L <= m) ans += Query(L,R,l,m,ls);if(R > m) ans += Query(L,R,m+1,r,rs);return ans;}int main(){while(~scanf("%d%d",&N,&Q)){for(int i = 1; i <= N; i ++)scanf("%lld",&d[i]);memset(sum,0,sizeof(sum));memset(add,0,sizeof(add));Build(1,N,1);char str[10];while(Q --){scanf("%s",str);int a,b,c;if(str[0] == 'C'){scanf("%d%d%d",&a,&b,&c);Update(a,b,c,1,N,1);}else{scanf("%d%d",&a,&b);ll ans = Query(a,b,1,N,1);printf("%lld\n",ans);}}}return 0;}