C – A Simple Problem with IntegersTime Limit:5000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64uSubmit Status Practice POJ 3468DescriptionYou have N integers, A1, A2, … , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.InputThe first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.The second line contains N numbers, the initial values of A1, A2, … , AN. -1000000000 ≤ Ai ≤ 1000000000.Each of the next Q lines represents an operation."C abc" means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000."Q ab" means querying the sum of Aa, Aa+1, … , Ab.OutputYou need to answer all Q commands in order. One answer in a line.Sample Input10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4Sample Output4559
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//线段树的区间更新，，用好延迟标记
//#include<bits/stdc++.h>#include<cstdio>#include<algorithm>using namespace std;const int maxn=500011;const int inf=999999999;#define lson (rt<<1),L,M#define rson (rt<<1|1),M+1,R#define M ((L+R)>>1)#define For(i,n) for(int i=0;i<(n);i++)template<class T>inline T read(T&x){char c;while((c=getchar())<=32);bool ok=false;if(c=='-')ok=true,c=getchar();for(x=0; c>32; c=getchar())x=x*10+c-'0';if(ok)x=-x;return x;}template<class T> inline void read_(T&x,T&y){read(x);read(y);}template<class T> inline void write(T x){if(x<0)putchar('-'),x=-x;if(x<10)putchar(x+'0');else write(x/10),putchar(x%10+'0');}template<class T>inline void writeln(T x){write(x);putchar('\n');}//——-IO template——typedef __int64 LL;struct node{LL sum;LL inv;node(){sum=0;inv=0;}}p[maxn<<2];void build(LL rt,LL L,LL R){if(L==R){scanf("%I64d",&p[rt].sum);//read(p[rt].sum);return;}build(lson);build(rson);p[rt].sum=p[rt<<1].sum+p[rt<<1|1].sum;}void update(LL rt,LL L,LL R,LL x,LL y,LL add){if(L==x&&R==y){p[rt].inv+=add;p[rt].sum+=(LL)add*(R-L+1);return;}if(p[rt].inv)///p[rt].inv可能收负数，之前写成了p[rt].inv>0 导致WA很多次{p[rt<<1].inv+=p[rt].inv;p[rt<<1|1].inv+=p[rt].inv;p[rt<<1].sum+=(LL)p[rt].inv*(M-L+1);p[rt<<1|1].sum+=(LL)p[rt].inv*(R-M);p[rt].inv=0;}if(y<=M)update(lson,x,y,add);else if(x>M)update(rson,x,y,add);else{update(lson,x,M,add);update(rson,M+1,y,add);}p[rt].sum=p[rt<<1].sum+p[rt<<1|1].sum;}LL query(LL rt,LL L,LL R,LL x,LL y){if(x==L&&y==R){return p[rt].sum;//+p[rt].inv*(R-L+1);}if(p[rt].inv){p[rt<<1].inv+=p[rt].inv;p[rt<<1|1].inv+=p[rt].inv;p[rt<<1].sum+=(LL)p[rt].inv*(M-L+1);p[rt<<1|1].sum+=(LL)p[rt].inv*(R-M);p[rt].inv=0;}if(y<=M)return query(lson,x,y);else if(x>M)return query(rson,x,y);else return query(lson,x,M)+query(rson,M+1,y);}int main(){//#ifndef ONLINE_JUDGE // freopen("in.txt","r",stdin); // #endif // ONLINE_JUDGELL m,n,i,j,k,t;while(~scanf("%I64d%I64d",&n,&m)){build(1,1,n);char op[2];while(m–){LL x,y;scanf("%s",op);scanf("%I64d%I64d",&x,&y);if(op[0]=='Q'){printf("%I64d\n",query(1,1,n,x,y));}else{LL z;scanf("%I64d",&z);update(1,1,n,x,y,z);}}}return 0;}
生活中若没有朋友，就像生活中没有阳光一样
