hdu多校第4场 B Harvest of Apples（莫队） _莫队

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Problem B. Harvest of ApplesTime Limit: 4000/2000 MS (java/Others)Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1600Accepted Submission(s): 604Problem Description There are n apples on a tree, numbered from 1 to n. Count the number of ways to pick at most m apples. Input The first line of the input contains an integer T (1≤T≤105) denoting the number of test cases. Each test case consists of one line with two integers n,m (1≤m≤n≤105). Output For each test case, print an integer representing the number of ways modulo 109+7. Sample Input 2 5 2 1000 500 Sample Output 16 924129523 Source 2018 Multi-University Training Contest 4 Recommend chendu|We have carefully selected several similar problems for you:6343 6342 6341 6340 6339

求C(n,0)+C(n,1)+C(n,2)+.....+C(n,m);
设S(n,m)=C(n,0)+C(n,1)+C(n,2)+.....+C(n,m);

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第一个式子易得，第二个式子：杨辉三角的 n,m=(n-1,m)+(n-1,m-1)
【hdu多校第4场 B Harvest of Apples（莫队）】那么就是这一行等于上一行的都用了2次，只有第最后一个用了一次
所以减去c（n-1，m）

#include< iostream> #include< stdio.h> #include< cmath> #include< algorithm> using namespace std; const int mod=1e9+7; #define ll long long const int maxn=1e5+7; ll jiecheng[maxn],inv[maxn]; ll ans[maxn]; int block; ll qsm(ll a,ll b) { ll ans=1; while(b){ if(b& 1) ans=ans*a%mod; a=a*a%mod; b> > =1; } return ans; } void init() { jiecheng[1] = 1; for(int i = 2; i < maxn; i++) jiecheng[i] = jiecheng[i-1] * i % mod; for(int i = 1; i < maxn; i++) inv[i] = qsm(jiecheng[i], mod-2); } struct node{ int l,r; int i; }modui[maxn]; bool cmp(node a,node b) { if(a.l/block==b.l/block) return a.r< b.r; return a.l< b.l; } ll C(ll n,ll m) {if(m == 0 || m == n) return 1; ll ans=1; ans=(jiecheng[n]*inv[m])%mod*inv[n-m]; ans=ans%mod; return ans; } int main() { init(); block = sqrt(maxn); int t; scanf("%d",& t); for(int i=0; i< t; i++) { scanf("%d%d",& modui[i].l,& modui[i].r); modui[i].i=i; } sort(modui,modui+t,cmp); int l=1,r=0; int sum=1; for(int i = 0; i < t; i++) { while(l < modui[i].l) sum = (2 * sum - C(l++, r) + mod) % mod; while(l > modui[i].l) sum = ((sum + C(--l, r))*inv[2]) % mod; while(r < modui[i].r) sum = (sum + C(l, ++r)) % mod; while(r > modui[i].r) sum = (sum - C(l, r--) + mod) % mod; ans[modui[i].i] = sum; } for(int i=0; i< t; i++) { printf("%lld\\n",ans[i]); }return 0; }