UVA 12532 Interval Product
线段树水题，忽略每个数的大小，只记住他们的正负即可，规规矩矩的代码。不过这是我第一次完全自己写的一次A的代码了。纪念一下。。。
#include <iostream> #include <cstdio> using namespace std; #define N 100010 int tree[4*N]; int x; char ans[100005]; void build(int l,int r,int rt) { if(l == r) { cin>>x; if(x>0) tree[rt] = 1; else if(x == 0) tree[rt] = 0; else tree[rt] = -1; return; } int mid = (l+r)/2; build(l,mid,2*rt); build(mid+1,r,2*rt+1); tree[rt] = tree[2*rt]*tree[2*rt+1]; } void change(int l,int r,int pos,int val,int rt) { if(l == r) { if(val<0) tree[rt] = -1; else if(val == 0) tree[rt] = 0; else tree[rt] = 1; return; } int mid = (l+r)/2; if(pos<=mid) change(l,mid,pos,val,2*rt); else change(mid+1,r,pos,val,2*rt+1); tree[rt] = tree[2*rt]*tree[2*rt+1]; } int query(int l,int r,int aa,int bb,int rt) { if(aa>r||bb<l) return 0; if(aa<=l&&bb>=r) return tree[rt]; int mid = (l+r)/2; if(bb<=mid) return query(l,mid,aa,bb,2*rt); if(aa>mid) return query(mid+1,r,aa,bb,2*rt+1); return query(l,mid,aa,bb,2*rt)*query(mid+1,r,aa,bb,2*rt+1); } int main() { int n,k,i,j; int ka; char ss[5]; int aa,bb; while(scanf("%d%d",&n,&k)!=EOF) { ka = 0; memset(ans,0,sizeof(ans)); build(1,n,1); for(i=0;i<k;i++) { scanf("%s %d %d",ss,&aa,&bb); if(ss[0] == 'C') { change(1,n,aa,bb,1); } else { int res = query(1,n,aa,bb,1); if(res > 0) ans[ka++] = '+'; else if(res < 0) ans[ka++] = '-'; else ans[ka++] = '0'; } } printf("%s\n",ans); } return 0; }